#include <bits/stdc++.h>
using namespace std;
void IOS() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll inf = 1e18;
ll powe(ll a, ll b, ll mod = MOD) {
  b %= (mod - 1);
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  IOS();
  ll tt = 1;
  cin >> tt;
  while (tt--) {
    ll n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    ll ans = 0;
    char la = '#', lb = '#';
    for (ll i = 0; i < n; i++) {
      if ((a[i] == '1' && b[i] == '0') || (a[i] == '0' && b[i] == '1')) {
        ans += 2;
        if (la == '0' && lb == '0') ans++;
        la = lb = '#';
      } else if (a[i] == '1' && b[i] == '1') {
        if (la == '0' && lb == '0')
          ans += 2, la = lb = '#';
        else
          la = lb = '1';
      } else {
        if (la == '1' && lb == '1')
          ans += 2, la = lb = '#';
        else if (la == '0' && lb == '0')
          ans++;
        else
          la = lb = '0';
      }
    }
    if (la == '0' && lb == '0') ans++;
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
void IOS() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll inf = 1e18;
ll powe(ll a, ll b, ll mod = MOD) {
  b %= (mod - 1);
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  IOS();
  ll tt = 1;
  cin >> tt;
  while (tt--) {
    ll n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    ll ans = 0;
    char la = '#', lb = '#';
    for (ll i = 0; i < n; i++) {
      if ((a[i] == '1' && b[i] == '0') || (a[i] == '0' && b[i] == '1')) {
        ans += 2;
        if (la == '0' && lb == '0') ans++;
        la = lb = '#';
      } else if (a[i] == '1' && b[i] == '1') {
        if (la == '0' && lb == '0')
          ans += 2, la = lb = '#';
        else
          la = lb = '1';
      } else {
        if (la == '1' && lb == '1')
          ans += 2, la = lb = '#';
        else if (la == '0' && lb == '0')
          ans++;
        else
          la = lb = '0';
      }
    }
    if (la == '0' && lb == '0') ans++;
    cout << ans << '\n';
  }
  return 0;
}

```