description: Stephen Queen wants to write a story. He is a very unusual writer, he uses only letters 'a', 'b', 'c', 'd' and 'e'!
  
  To compose a story, Stephen wrote out n words consisting of the first 5 lowercase letters of the Latin alphabet. He wants to select the maximum number of words to make an interesting story.
  
  Let a story be a sequence of words that are not necessarily different. A story is called interesting if there exists a letter which occurs among all words of the story more times than all other letters together.
  
  For example, the story consisting of three words "bac", "aaada", "e" is interesting (the letter 'a' occurs 5 times, all other letters occur 4 times in total). But the story consisting of two words "aba", "abcde" is not (no such letter that it occurs more than all other letters in total).
  
  You are given a sequence of n words consisting of letters 'a', 'b', 'c', 'd' and 'e'. Your task is to choose the maximum number of them to make an interesting story. If there's no way to make a non-empty story, output 0.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of the words in the sequence. Then n lines follow, each of them contains a word — a non-empty string consisting of lowercase letters of the Latin alphabet. The words in the sequence may be non-distinct (i. e. duplicates are allowed). Only the letters 'a', 'b', 'c', 'd' and 'e' may occur in the words.
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5; the sum of the lengths of all words over all test cases doesn't exceed 4 ⋅ 10^5.
  
  Output
  
  For each test case, output the maximum number of words that compose an interesting story. Print 0 if there's no way to make a non-empty interesting story.
  
  Example
  
  Input
  
  
  6
  3
  bac
  aaada
  e
  3
  aba
  abcde
  aba
  2
  baba
  baba
  4
  ab
  ab
  c
  bc
  5
  cbdca
  d
  a
  d
  e
  3
  b
  c
  ca
  
  
  Output
  
  
  3
  2
  0
  2
  3
  2
  
  Note
  
  In the first test case of the example, all 3 words can be used to make an interesting story. The interesting story is "bac aaada e".
  
  In the second test case of the example, the 1-st and the 3-rd words can be used to make an interesting story. The interesting story is "aba aba". Stephen can't use all three words at the same time.
  
  In the third test case of the example, Stephen can't make a non-empty interesting story. So the answer is 0.
  
  In the fourth test case of the example, Stephen can use the 3-rd and the 4-th words to make an interesting story. The interesting story is "c bc".
solution: import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  
  public class cf2 {
      static PrintWriter out;
      static int MOD = 1000000007;
      static FastReader scan;
  
      /*-------- I/O using short named function ---------*/
      public static String ns() {
          return scan.next();
      }
  
      public static int ni() {
          return scan.nextInt();
      }
  
      public static long nl() {
          return scan.nextLong();
      }
  
      public static double nd() {
          return scan.nextDouble();
      }
  
      public static String nln() {
          return scan.nextLine();
      }
  
      public static void p(Object o) {
          out.print(o);
      }
  
      public static void ps(Object o) {
          out.print(o + " ");
      }
  
      public static void pn(Object o) {
          out.println(o);
      }
  
      /*-------- for output of an array ---------------------*/
      static void iPA(int arr[]) {
          StringBuilder output = new StringBuilder();
          for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
          out.println(output);
      }
  
      static void lPA(long arr[]) {
          StringBuilder output = new StringBuilder();
          for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
          out.println(output);
      }
  
      static void sPA(String arr[]) {
          StringBuilder output = new StringBuilder();
          for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
          out.println(output);
      }
  
      static void dPA(double arr[]) {
          StringBuilder output = new StringBuilder();
          for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
          out.println(output);
      }
  
      /*-------------- for input in an array ---------------------*/
      static void iIA(int arr[]) {
          for (int i = 0; i < arr.length; i++) arr[i] = ni();
      }
  
      static void lIA(long arr[]) {
          for (int i = 0; i < arr.length; i++) arr[i] = nl();
      }
  
      static void sIA(String arr[]) {
          for (int i = 0; i < arr.length; i++) arr[i] = ns();
      }
  
      static void dIA(double arr[]) {
          for (int i = 0; i < arr.length; i++) arr[i] = nd();
      }
  
      /*------------ for taking input faster ----------------*/
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static ArrayList<Integer> sieveOfEratosthenes(int n) {
          // Create a boolean array
          // "prime[0..n]" and
          // initialize all entries
          // it as true. A value in
          // prime[i] will finally be
          // false if i is Not a
          // prime, else true.
          boolean prime[] = new boolean[n + 1];
          for (int i = 0; i <= n; i++)
              prime[i] = true;
  
          for (int p = 2; p * p <= n; p++) {
              // If prime[p] is not changed, then it is a
              // prime
              if (prime[p] == true) {
                  // Update all multiples of p
                  for (int i = p * p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          ArrayList<Integer> arr = new ArrayList<>();
          // Print all prime numbers
          for (int i = 2; i <= n; i++) {
              if (prime[i] == true)
                  arr.add(i);
          }
          return arr;
      }
  
      // Method to check if x is power of 2
      static boolean isPowerOfTwo(int x) {
          return x != 0 && ((x & (x - 1)) == 0);
      }
  
      //Method to return lcm of two numbers
      static int gcd(int a, int b) {
          return a == 0 ? b : gcd(b % a, a);
      }
  
      //Method to count digit of a number
      static int countDigit(long n) {
          String sex = Long.toString(n);
          return sex.length();
      }
  
      static void reverse(int a[]) {
          int i, k, t;
          int n = a.length;
          for (i = 0; i < n / 2; i++) {
              t = a[i];
              a[i] = a[n - i - 1];
              a[n - i - 1] = t;
          }
      }
      static long nCr(long n, long r)
      {
          return fact(n) / (fact(r) *
                  fact(n - r));
      }
  
      // Returns factorial of n
      static long fact(long n)
      {
          long res = 1;
          for (long i = 2; i <= n; i++)
              res = res * i;
          return res;
      }
      // Get all prime numbers till N using seive of Eratosthenes
      public static List<Integer> getPrimesTill(int n){
          boolean[] arr = new boolean[n+1];
          List<Integer> primes = new LinkedList<>();
          for(int i=2; i<=n; i++){
              if(!arr[i]){
                  primes.add(i);
                  for(long j=(i*i); j<=n; j+=i){
                      arr[(int)j] = true;
                  }
              }
          }
          return primes;
      }
  
      static final Random random = new Random();
  
      //Method for sorting
      static void ruffleSort(int[] arr) {
          int n = arr.length;
          for (int i = 0; i < n; i++) {
              int j = random.nextInt(n);
              int temp = arr[j];
              arr[j] = arr[i];
              arr[i] = temp;
          }
          Arrays.sort(arr);
      }
  
      static void sortadv(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long value : a) {
              l.add(value);
          }
          Collections.sort(l);
          for (int i = 0; i < l.size(); i++)
              a[i] = l.get(i);
      }
  
      //Method for checking if a number is prime or not
      static boolean isPrime(int n) {
          if (n <= 1) return false;
          if (n <= 3) return true;
          if (n % 2 == 0 || n % 3 == 0) return false;
          for (int i = 5; i * i <= n; i = i + 6)
              if (n % i == 0 || n % (i + 2) == 0)
                  return false;
          return true;
      }
  
      public static void main(String[] args) throws java.lang.Exception {
          OutputStream outputStream = System.out;
          out = new PrintWriter(outputStream);
          scan = new FastReader();
          //for fast output sometimes
          StringBuilder sb = new StringBuilder();
          // prefix sum/dp , odd-even segregate , *brute force , 2 pointer , pattern
          // but first observe!
          int t = ni();
          while (t-- != 0) {
              int n=ni();
              String[] s=new String[n];
              sIA(s);
              int[] a=new int[n];
              int[] b=new int[n];
              int[] c=new int[n];
              int[] d=new int[n];
              int[] e=new int[n];
              for(int i=0;i<n;i++){
                  for(char ch:s[i].toCharArray())
                  {
                      if(ch=='a')
                          a[i]++;
                      if(ch=='b')
                          b[i]++;
                      if(ch=='c')
                          c[i]++;
                      if(ch=='d')
                          d[i]++;
                      if(ch=='e')
                          e[i]++;
                  }
                  a[i]=2*a[i]-s[i].length();
                  b[i]=2*b[i]-s[i].length();
                  c[i]=2*c[i]-s[i].length();
                  d[i]=2*d[i]-s[i].length();
                  e[i]=2*e[i]-s[i].length();
              }
              ruffleSort(a);reverse(a);
              ruffleSort(b);reverse(b);
              ruffleSort(c);reverse(c);
              ruffleSort(d);reverse(d);
              ruffleSort(e);reverse(e);int max=0;
              int count=0;int
                      sum=0;
              for (int i=0;i<n;i++){
                  if(sum+a[i]>0){
                      sum+=a[i];
                      count++;
                  }
              }
              //iPA(a);
              sum=0;
              if(count>max)
                  max=count;
              count=0;
              for (int i=0;i<n;i++){
                  if(sum+b[i]>0){
                      sum+=b[i];
                      count++;
                  }
              }
              sum=0;
              if(count>max)
                  max=count;
              count=0;
              for (int i=0;i<n;i++){
                  if(sum+c[i]>0){
                      sum+=c[i];
                      count++;
                  }
              }
              sum=0;
              if(count>max)
                  max=count;
              count=0;
              for (int i=0;i<n;i++){
                  if(sum+d[i]>0){
                      sum+=d[i];
                      count++;
                  }
              }
              sum=0;
              if(count>max)
                  max=count;
              count=0;
              for (int i=0;i<n;i++){
                  if(sum+e[i]>0){
                      sum+=e[i];
                      count++;
                  }
              }
              if(count>max)
                  max=count;
              pn(max);
          }
          out.flush();
          out.close();
      }
  }
  