description: There is a chessboard of size n by n. The square in the i-th row from top and j-th column from the left is labelled (i,j).
  
  Currently, Gregor has some pawns in the n-th row. There are also enemy pawns in the 1-st row. On one turn, Gregor moves one of his pawns. A pawn can move one square up (from (i,j) to (i-1,j)) if there is no pawn in the destination square. Additionally, a pawn can move one square diagonally up (from (i,j) to either (i-1,j-1) or (i-1,j+1)) if and only if there is an enemy pawn in that square. The enemy pawn is also removed.
  
  Gregor wants to know what is the maximum number of his pawns that can reach row 1?
  
  Note that only Gregor takes turns in this game, and the enemy pawns never move. Also, when Gregor's pawn reaches row 1, it is stuck and cannot make any further moves.
  
  Input
  
  The first line of the input contains one integer t (1≤ t≤ 2⋅ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of three lines. The first line contains a single integer n (2≤ n≤ 2⋅{10}^{5}) — the size of the chessboard.
  
  The second line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to an enemy pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  The third line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to a Gregor's pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  It is guaranteed that the sum of n across all test cases is less than 2⋅{10}^{5}.
  
  Output
  
  For each test case, print one integer: the maximum number of Gregor's pawns which can reach the 1-st row.
  
  Example
  
  Input
  
  
  4
  3
  000
  111
  4
  1111
  1111
  3
  010
  010
  5
  11001
  00000
  
  
  Output
  
  
  3
  4
  0
  0
  
  Note
  
  In the first example, Gregor can simply advance all 3 of his pawns forward. Thus, the answer is 3.
  
  In the second example, Gregor can guarantee that all 4 of his pawns reach the enemy row, by following the colored paths as demonstrated in the diagram below. Remember, only Gregor takes turns in this "game"!
  
  <image>
  
  In the third example, Gregor's only pawn is stuck behind the enemy pawn, and cannot reach the end.
  
  In the fourth example, Gregor has no pawns, so the answer is clearly 0.
solution:  //package eround101;
  
  import java.util.*;
  import java.io.*;
  import java.lang.*;
  import java.util.StringTokenizer;
  
  public class C101 {
  
      static HritikScanner sc = new HritikScanner();
      static PrintWriter pw = new PrintWriter(System.out, true);
      final static int MOD = (int) 1e9 + 7;
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) {
  
          int t = ni();
          while (t-- > 0) {
              solve();
          }
      }
  
      static void solve() {
          int n= ni();
          char[] e = sc.next().toCharArray();
          char[] m = sc.next().toCharArray();
          int c = 0;
          if(m[0] == '1')
          {
              if(e[0] == '0')c++;
              else if(e[1] == '1')
              {
                  c++;
                  e[1] = 'a';
              }
          }
          if(m[n-1] == '1')
          {
              if(e[n-1] == '0')c++;
              else if(e[n-2] == '1')
              {
                  c++;
                  e[n-2] = 'a';
              }
          }
          for(int i = 1; i < n-1; i++)
          {
              if(m[i] == '1')
              {
                  if(e[i] == '0')
                  {
                      c++;
                  }
                  else if(e[i-1] == '1')
                  {
                      c++;
                      e[i-1] = 'a';
                  }
                  else if(e[i+1] == '1')
                  {
                      c++;
                      e[i+1] = 'a';
                  }
              }
          }
          pl(c);
      }
      
      
      
      
      /////////////////////////////////////////////////////////////////////////////////
      static int[] nextIntArray(int n) {
          int[] arr = new int[n];
          int i = 0;
          while (i < n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      static long[] nextLongArray(int n) {
          long[] arr = new long[n];
          int i = 0;
          while (i < n) {
              arr[i++] = nl();
          }
          return arr;
      }
  
      static int[] nextIntArray1(int n) {
          int[] arr = new int[n + 1];
          int i = 1;
          while (i <= n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static int ni() {
          return sc.nextInt();
      }
  
      static long nl() {
          return sc.nextLong();
      }
  
      static double nd() {
          return sc.nextDouble();
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static void pl() {
          pw.println();
      }
  
      static void p(Object o) {
          pw.print(o+ " ");
      }
  
      static void pl(Object o) {
          pw.println(o);
      }
  
      static void psb(StringBuilder sb) {
          pw.print(sb);
      }
  
      static void pa(Object arr[]) {
          for (Object o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(int arr[]) {
          for (int o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(long arr[]) {
          for (long o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(double arr[]) {
          for (double o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(char arr[]) {
          for (char o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(List list) {
          for (Object o : list) {
              p(o);
          }
          pl();
      }
  
      static void pa(Object[][] arr) {
  
          for (int i = 0; i < arr.length; ++i) {
              for (Object o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(int[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (int o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(long[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (long o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(char[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (char o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(double[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (double o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
  /////////////////////////////////////////////////////////////////////////////////
      static void print(Object s) {
          System.out.println(s);
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      //-----------HritikScanner class for faster input----------//
      static class HritikScanner {
  
          BufferedReader br;
          StringTokenizer st;
  
          public HritikScanner() {
              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 class Pair implements Comparable<Pair> {
  
          int num, row,col;
  
          Pair(int num, int row, int col) {
              this.num = num;
              this.row = row;
              this.col = col;
          }
  
          public int get1() {
              return num;
          }
  
          public int get2() {
              return row;
          }
  
          public int compareTo(Pair A) {
              return this.num - A.num;
          }
          
          public String toString()
          {
              return num +" "+ row+" "+col;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      // Function to return gcd of a and b time complexity O(log(a+b))
      static long gcd(long a, long b) {
          if (a == 0) {
              return b;
          }
          return gcd(b % a, a);
      }
  
      // method to return LCM of two numbers 
      static long lcm(int a, int b) {
          return (a / gcd(a, b)) * b;
      }
  
      //////////////////////////////////////////////////////////////////
      static boolean isPrime(long n) {
          // Corner cases 
          if (n <= 1) {
              return false;
          }
          if (n <= 3) {
              return true;
          }
  
          // This is checked so that we can skip  
          // middle five numbers in below loop 
          if (n % 2 == 0 || n % 3 == 0) {
              return false;
          }
  
          for (long i = 5; i * i <= n; i = i + 6) {
              if (n % i == 0 || n % (i + 2) == 0) {
                  return false;
              }
          }
  
          return true;
      }
      //////////////////////////////////////////////////////////////////
  
      static boolean isPowerOfTwo(long n) {
          if (n == 0) {
              return false;
          }
  
          return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                  == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
      }
  
      public static long getFact(int n) {
          long ans = 1;
          while (n > 0) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          return ans;
      }
  
      public static long pow(long n, int pow) {
          if (pow == 0) {
              return 1;
          }
          long temp = pow(n, pow / 2) % MOD;
          temp *= temp;
          temp %= MOD;
          if (pow % 2 == 1) {
              temp *= n;
          }
          temp %= MOD;
          return temp;
  
      }
  
      public static long nCr(int n, int r) {
          long ans = 1;
          int temp = n - r;
          while (n > temp) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
          ans %= MOD;
          return ans;
      }
  
      //////////////////////////////////////////////////////////////////
      // method returns Nth power of A 
      static double nthRoot(int A, int N) {
  
          // intially guessing a random number between 
          // 0 and 9 
          double xPre = Math.random() % 10;
  
          // smaller eps, denotes more accuracy 
          double eps = 0.001;
  
          // initializing difference between two 
          // roots by INT_MAX 
          double delX = 2147483647;
  
          // xK denotes current value of x 
          double xK = 0.0;
  
          // loop untill we reach desired accuracy 
          while (delX > eps) {
              // calculating current value from previous 
              // value by newton's method 
              xK = ((N - 1.0) * xPre
                      + (double) A / Math.pow(xPre, N - 1)) / (double) N;
              delX = Math.abs(xK - xPre);
              xPre = xK;
          }
  
          return xK;
      }
  
  }
  