A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.

Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.

You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.

What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.

Each of the next two lines contains a binary string of length n — the rows of the bi-table.

It's guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.

Example

Input


4
7
0101000
1101100
5
01100
10101
2
01
01
6
000000
111111


Output


8
8
2
12

Note

In the first test case you can cut the bi-table as follows:

  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.



The sum of \operatorname{MEX} is 8.
#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5;
int t, n, chk[maxn], s[maxn];
string s1, s2;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> s1;
    cin >> s2;
    int cnt1 = 0;
    int cnt0 = 0;
    int cnt10 = 0;
    for (int i = 0; i < n; ++i) {
      chk[i] = 0;
      if (s1[i] == '0' && s2[i] == '0')
        ++cnt0;
      else {
        if (s1[i] == '1' && s2[i] == '1')
          ++cnt1;
        else
          ++cnt10;
      }
    }
    if (!cnt10) {
      if (!cnt1)
        cout << n << '\n';
      else {
        if (!cnt0)
          cout << 0 << '\n';
        else {
          int ans = 0;
          for (int i = 0; i < n; ++i) {
            if (chk[i]) continue;
            chk[i] = 1;
            if (s1[i] == '0') {
              if (s1[i - 1] == '1' && chk[i - 1] == 0) {
                chk[i - 1] = 1;
                ans += 2;
              } else {
                if (s1[i + 1] == '1' && chk[i + 1] == 0) {
                  ans += 2;
                  chk[i + 1] = 1;
                } else
                  ans += 1;
              }
            }
            if (s1[i] == '1') {
              if (s1[i - 1] == '0' && chk[i - 1] == 0) {
                chk[i - 1] = 1;
                ans += 2;
              } else {
                if (s1[i + 1] == '0' && chk[i + 1] == 0) {
                  ans += 2;
                  chk[i + 1] = 1;
                }
              }
            }
          }
          cout << ans << '\n';
        }
      }
      continue;
    }
    for (int i = 0; i < n; ++i) {
      chk[i] = 0;
      if (s1[i] == '0' && s2[i] == '0')
        s[i] = 1;
      else {
        if (s1[i] == '1' && s2[i] == '1')
          s[i] = 0;
        else
          s[i] = 2;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (chk[i]) continue;
      chk[i] = 1;
      if (s[i] == 2) {
        ans += 2;
        continue;
      }
      if (s[i] == 1) {
        if (s[i - 1] == 0 && chk[i - 1] == 0) {
          chk[i - 1] = 1;
          ans += 2;
        } else {
          if (s[i + 1] == 0 && chk[i + 1] == 0) {
            ans += 2;
            chk[i + 1] = 1;
          } else
            ans += 1;
        }
      }
      if (s[i] == 0) {
        if (s[i - 1] != 0 && chk[i - 1] == 0) {
          chk[i - 1] = 1;
          ans += 2;
        } else {
          if (s[i + 1] != 0 && chk[i + 1] == 0) {
            ans += 2;
            chk[i + 1] = 1;
          }
        }
      }
    }
    cout << ans << '\n';
  }
}
