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 = 1e5 + 10;
int T, n, cnt;
char s[3][maxn];
bool flag[maxn];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    scanf("%s", s[1] + 1);
    scanf("%s", s[2] + 1);
    cnt = 0;
    for (int i = 1; i <= n; i++) flag[i] = 0;
    for (int i = 1; i <= n; i++)
      if (s[1][i] - '0' + s[2][i] - '0' == 1) flag[i] = 1, cnt += 2;
    for (int i = 1; i <= n; i++) {
      if (s[1][i] - '0' + s[2][i] - '0' == 2) {
        bool check = 0;
        for (int j = i - 1; j >= 1; j--) {
          if (s[1][j] - '0' + s[2][j] - '0' >= 1)
            break;
          else if (flag[j] == 0) {
            flag[j] = 1;
            cnt += 2;
            check = 1;
          }
        }
        if (!check) {
          for (int j = i + 1; j <= n; j++) {
            if (s[1][j] - '0' + s[2][j] - '0' >= 1)
              break;
            else if (flag[j] == 0) {
              flag[j] = 1;
              cnt += 2;
            }
          }
        }
      }
    }
    for (int i = 1; i <= n; i++)
      if (s[1][i] - '0' + s[2][i] - '0' == 0 && flag[i] == 0) cnt++;
    printf("%d\n", cnt);
  }
}
