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: #include <bits/stdc++.h>
  using namespace std;
  bool isPrime(long long n) {
    if (n == 1) {
      return false;
    } else if (n == 2 || n == 3) {
      return true;
    } else {
      for (long long i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
          return false;
        }
      }
    }
    return true;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) {
      return b;
    } else {
      return gcd(b % a, a);
    }
  }
  long long lcm(long long a, long long b) {
    return (max(a, b) / gcd(a, b)) * min(a, b);
  }
  bool isEven(long long a) {
    if (a % 2 == 0) {
      return true;
    }
    return false;
  }
  bool isSquare(long long x) {
    long long n1 = sqrt(x);
    double n2 = sqrt(x);
    if (n2 - n1 == 0) {
      return true;
    }
    return false;
  }
  bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
  void solve(void) {
    long long t;
    cin >> t;
    while (t--) {
      long long x;
      cin >> x;
      if (x == 5) {
        cout << 2 << " " << 4;
      } else {
        cout << 2 << " " << x / 2;
      }
      cout << endl;
    }
  }
  void solve1(void) {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      string s1, s2;
      cin >> s1 >> s2;
      long long cnt = 0;
      for (long long i = 0; i < n; i++) {
        if (s2[i] == '1') {
          if (i == 0) {
            if (s1[i] == '0') {
              s1[i] = '1';
              s2[i] = '0';
              cnt++;
            } else {
              if (s1[i + 1] == '1' && i + 1 < n) {
                s1[i + 1] = '1';
                s2[i] = '0';
                cnt++;
              }
            }
          } else {
            if (s1[i - 1] == '1') {
              s1[i - 1] = '1';
              s2[i] = '0';
              cnt++;
            } else if (s1[i] == '0') {
              s1[i] = '1';
              s2[i] = '0';
              cnt++;
            } else {
              if (s1[i + 1] == '1' && i + 1 < n) {
                s1[i + 1] = '1';
                s2[i] = '0';
                cnt++;
              }
            }
          }
        }
      }
      cout << cnt << endl;
    }
  }
  void solve2(void) {
    long long t;
    cin >> t;
    while (t--) {
      long long n, count = 0;
      cin >> n;
      bool str[n];
      for (int i = 0; i < n; i++) {
        str[i] = false;
      }
      string gre;
      string enemy;
      cin >> enemy;
      cin >> gre;
      for (int i = 0; i < gre.size(); i++) {
        if (gre[i] == '1') {
          if (enemy[i] == '0' && str[i] == false) {
            count++;
            str[i] = true;
          } else if (enemy[i] == '1' && i - 1 >= 0 && enemy[i - 1] == '1' &&
                     str[i - 1] == false) {
            count++;
            str[i - 1] = true;
          } else if (enemy[i] == '1' && i + 1 < n && enemy[i + 1] == '1' &&
                     str[i + 1] == false) {
            count++;
            str[i + 1] = true;
          }
        }
      }
      cout << count << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    solve2();
  }
  