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 mod = 1e9 + 7;
int t, n;
long long cnt, ans;
string s, c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> s >> c;
    n = s.size();
    ans = cnt = 0;
    bool flag = 0, ff1 = 1, ff2 = 1, f0 = 1, f00 = 1;
    for (int i = 0; i < n; i++) {
      if (s[i] != c[i])
        ans += 2, flag = 0;
      else if (s[i] == '1' && c[i] == '1') {
        flag = 1;
      } else {
        if (flag)
          ans += 2, flag = 0;
        else
          ans++;
      }
    }
    flag = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] != c[i])
        cnt += 2, flag = 0;
      else if (s[i] == '1' && c[i] == '1') {
        flag = 1;
      } else {
        if (flag)
          cnt += 2, flag = 0;
        else
          cnt++;
      }
    }
    int i = 0, j = n - 1;
    while (s[i] == '1' && i < n) i++;
    while (s[j] == '1' && j >= 0) j--;
    if (!i && j == n - 1 || i == n && j < 0) ff1 = 0;
    for (i; i <= j; i++)
      if (s[i] == '1') {
        ff1 = 0;
        break;
      }
    i = 0, j = n - 1;
    while (c[i] == '1' && i < n) i++;
    while (c[j] == '1' && j >= 0) j--;
    if ((!i && j == n - 1) || i == n && j < 0) ff2 = 0;
    for (i; i <= j; i++)
      if (c[i] == '1') {
        ff2 = 0;
        break;
      }
    i = 0, j = n - 1;
    while (s[i] == '0' && i < n) i++;
    while (s[j] == '0' && j >= 0) j--;
    if (!i && j == n - 1 || i == n && j < 0) f0 = 0;
    for (i; i <= j; i++)
      if (s[i] == '0') {
        f0 = 0;
        break;
      }
    i = 0, j = n - 1;
    while (c[i] == '0' && i < n) i++;
    while (c[j] == '0' && j >= 0) j--;
    if (!i && j == n - 1 || i == n && j < 0) f00 = 0;
    for (i; i <= j; i++)
      if (c[i] == '0') {
        f00 = 0;
        break;
      }
    if ((ff1 && ff2) || (f0 && f00)) ans++;
    cout << max(ans, cnt) << endl;
  }
  return 0;
}
