#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
int mex(int x, int y) {
  if (x != y) return 2;
  return x ^ 1;
}
void solve() {
  int n;
  cin >> n;
  array<string, 2> s;
  cin >> s[0] >> s[1];
  vector<int> dps(n + 1, 0);
  array<array<int, 2>, 2> last;
  last[0][0] = last[0][1] = last[1][0] = last[1][1] = -1;
  for (int i = 0; i < n; ++i) {
    int& dp = dps[i + 1];
    int x = s[0][i] - '0';
    int y = s[1][i] - '0';
    assert(0 <= x && x <= 1);
    assert(0 <= y && y <= 1);
    dp = dps[i] + mex(x, y);
    if (last[0][x ^ 1] >= 0) {
      dp = max(dp, dps[last[0][x ^ 1]] + 2);
    }
    if (last[0][y ^ 1] >= 0) {
      dp = max(dp, dps[last[0][y ^ 1]] + 2);
    }
    if (last[1][x ^ 1] >= 0) {
      dp = max(dp, dps[last[1][x ^ 1]] + 2);
    }
    if (last[1][y ^ 1] >= 0) {
      dp = max(dp, dps[last[1][y ^ 1]] + 2);
    }
    last[0][x] = i;
    last[1][y] = i;
  }
  cout << dps[n] << '\n';
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int T;
  cin >> T;
  while (T-- > 0) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
int mex(int x, int y) {
  if (x != y) return 2;
  return x ^ 1;
}
void solve() {
  int n;
  cin >> n;
  array<string, 2> s;
  cin >> s[0] >> s[1];
  vector<int> dps(n + 1, 0);
  array<array<int, 2>, 2> last;
  last[0][0] = last[0][1] = last[1][0] = last[1][1] = -1;
  for (int i = 0; i < n; ++i) {
    int& dp = dps[i + 1];
    int x = s[0][i] - '0';
    int y = s[1][i] - '0';
    assert(0 <= x && x <= 1);
    assert(0 <= y && y <= 1);
    dp = dps[i] + mex(x, y);
    if (last[0][x ^ 1] >= 0) {
      dp = max(dp, dps[last[0][x ^ 1]] + 2);
    }
    if (last[0][y ^ 1] >= 0) {
      dp = max(dp, dps[last[0][y ^ 1]] + 2);
    }
    if (last[1][x ^ 1] >= 0) {
      dp = max(dp, dps[last[1][x ^ 1]] + 2);
    }
    if (last[1][y ^ 1] >= 0) {
      dp = max(dp, dps[last[1][y ^ 1]] + 2);
    }
    last[0][x] = i;
    last[1][y] = i;
  }
  cout << dps[n] << '\n';
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int T;
  cin >> T;
  while (T-- > 0) {
    solve();
  }
  return 0;
}

```