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: //Utilities
  import java.io.*;
  import java.util.*;
  
  public class D {
  	static int t;
  	static int n;
  	static char[] ch;
  	static int[][] cnt;
  	static int[] len;
  	static int max, cmp;
  	static int net, cur;
  	
  	public static void main(String[] args) throws IOException {
  		t = in.iscan();
  		while (t-- > 0) {
  			n = in.iscan(); cnt = new int[n][26]; len = new int[n];
  			for (int i = 0; i < n; i++) {
  				ch = in.sscan().toCharArray();
  				len[i] = ch.length;
  				for (int j = 0; j < ch.length; j++) {
  					cnt[i][ch[j]-'a']++;
  				}
  			}
  			max = 0;
  			for (int letter = 0; letter < 26; letter++) {
  				ArrayList<Integer> tmp = new ArrayList<Integer>();
  				for (int i = 0; i < n; i++) {
  					net = cnt[i][letter] - (len[i] - cnt[i][letter]);
  					tmp.add(net);
  				}
  				Collections.sort(tmp, Collections.reverseOrder());
  				cmp = 0; cur = 0;
  				for (int i = 0; i < tmp.size(); i++) {
  					if (tmp.get(i) > 0) {
  						cur += tmp.get(i);
  						cmp++;
  					}
  					else {
  						if (cur + tmp.get(i) > 0) {
  							cur += tmp.get(i);
  							cmp++;
  						}
  						else {
  							break;
  						}
  					}
  				}
  				max = Math.max(max, cmp);
  			}
  			out.println(max);
  		}
  		out.close();
  	} 
  	
  	static INPUT in = new INPUT(System.in);
  	static PrintWriter out = new PrintWriter(System.out);
  	private static class INPUT {
  
  		private InputStream stream;
  		private byte[] buf = new byte[1024];
  		private int curChar, numChars;
  
  		public INPUT (InputStream stream) {
  			this.stream = stream;
  		}
  
  		public INPUT (String file) throws IOException {
  			this.stream = new FileInputStream (file);
  		}
  
  		public int cscan () throws IOException {
  			if (curChar >= numChars) {
  				curChar = 0;
  				numChars = stream.read (buf);
  			}
  			
  			if (numChars == -1)
  				return numChars;
  
  			return buf[curChar++];
  		}
  
  		public int iscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			int res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public String sscan () throws IOException {
  			int c = cscan ();
  			
  			while (space (c))
  				c = cscan ();
  
  			StringBuilder res = new StringBuilder ();
  
  			do {
  				res.appendCodePoint (c);
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res.toString ();
  		}
  
  		public double dscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			double res = 0;
  
  			while (!space (c) && c != '.') {
  				if (c == 'e' || c == 'E')
  					return res * UTILITIES.fast_pow (10, iscan ());
  				
  				res *= 10;
  				res += c - '0';
  				c = cscan ();
  			}
  
  			if (c == '.') {
  				c = cscan ();
  				double m = 1;
  
  				while (!space (c)) {
  					if (c == 'e' || c == 'E')
  						return res * UTILITIES.fast_pow (10, iscan ());
  
  					m /= 10;
  					res += (c - '0') * m;
  					c = cscan ();
  				}
  			}
  
  			return res * sgn;
  		}
  
  		public long lscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			long res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public boolean space (int c) {
  			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
  		}
  	}
  
  	public static class UTILITIES {
  
  		static final double EPS = 10e-6;
  
  		public static int lower_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] >= x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static int upper_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] > x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static long gcd (long a, long b) {
  			return b == 0 ? a : gcd (b, a % b);
  		}
  
  		public static long lcm (long a, long b) {
  			return a * b / gcd (a, b);
  		}
  
  		public static long fast_pow_mod (long b, long x, int mod) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  
  			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  		}
  
  		public static int fast_pow (int b, int x) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow (b * b, x / 2);
  
  			return b * fast_pow (b * b, x / 2);
  		}
  
  		public static long choose (long n, long k) {
  			k = Math.min (k, n - k);
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = val * (n - i) / (i + 1);
  
  			return val;
  		}
  
  		public static long permute (int n, int k) {
  			if (n < k) return 0;
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = (val * (n - i));
  
  			return val;
  		}
  		
  		// start of permutation and lower/upper bound template
  		public static void nextPermutation(int[] nums) {
  		    //find first decreasing digit
  		    int mark = -1;
  		    for (int i = nums.length - 1; i > 0; i--) {
  		        if (nums[i] > nums[i - 1]) {
  		            mark = i - 1;
  		            break;
  		        }
  		    }
  		 
  		    if (mark == -1) {
  		        reverse(nums, 0, nums.length - 1);
  		        return;
  		    }
  		 
  		    int idx = nums.length-1;
  		    for (int i = nums.length-1; i >= mark+1; i--) {
  		        if (nums[i] > nums[mark]) {
  		            idx = i;
  		            break;
  		        }
  		    }
  		 
  		    swap(nums, mark, idx);
  		 
  		    reverse(nums, mark + 1, nums.length - 1);
  		}
  		 
  		public static void swap(int[] nums, int i, int j) {
  		    int t = nums[i];
  		    nums[i] = nums[j];
  		    nums[j] = t;
  		}
  		 
  		public static void reverse(int[] nums, int i, int j) {
  		    while (i < j) {
  		        swap(nums, i, j);
  		        i++;
  		        j--;
  		    }
  		}
  		
  		static int lower_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] >= cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  	 
  		static int upper_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] > cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  		// end of permutation and lower/upper bound template
  	}
  }
  