#include <bits/stdc++.h>
using namespace std;
int MOD = 1e9 + 7;
int OO = 1e8;
const int N = 1e5 + 10, M = N << 1;
int n, m, k, T;
int a, b, c;
string str, s;
inline int getMX(int msk) {
  for (int i = 0; i < 3; ++i) {
    if ((msk >> i) & 1) continue;
    return i;
  }
  return 0;
}
inline void SetMsk(int &msk, char x) { msk |= (1 << (x - '0')); }
int dp[N][4], vis[N][4], vid;
int Solve(int idx, int msk) {
  if (idx == n) return getMX(msk);
  int &ret = dp[idx][msk];
  if (vis[idx][msk] == vid) return ret;
  vis[idx][msk] = vid;
  SetMsk(msk, str[idx]);
  SetMsk(msk, s[idx]);
  ret = Solve(idx + 1, msk);
  ret = max(ret, getMX(msk) + Solve(idx + 1, 0));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL), cerr.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> str >> s;
    ++vid;
    cout << Solve(0, 0) << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
int MOD = 1e9 + 7;
int OO = 1e8;
const int N = 1e5 + 10, M = N << 1;
int n, m, k, T;
int a, b, c;
string str, s;
inline int getMX(int msk) {
  for (int i = 0; i < 3; ++i) {
    if ((msk >> i) & 1) continue;
    return i;
  }
  return 0;
}
inline void SetMsk(int &msk, char x) { msk |= (1 << (x - '0')); }
int dp[N][4], vis[N][4], vid;
int Solve(int idx, int msk) {
  if (idx == n) return getMX(msk);
  int &ret = dp[idx][msk];
  if (vis[idx][msk] == vid) return ret;
  vis[idx][msk] = vid;
  SetMsk(msk, str[idx]);
  SetMsk(msk, s[idx]);
  ret = Solve(idx + 1, msk);
  ret = max(ret, getMX(msk) + Solve(idx + 1, 0));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL), cerr.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> str >> s;
    ++vid;
    cout << Solve(0, 0) << "\n";
  }
  return 0;
}

```