#include <bits/stdc++.h>
long long q_pow(long long x, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
using namespace std;
const long long maxn = 1e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
char s1[maxn];
char s2[maxn];
void solve() {
  int n;
  cin >> n;
  cin >> (s1 + 1);
  cin >> (s2 + 1);
  int ans = 0;
  int cnt0 = 0, cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    s1[i] -= '0';
    s2[i] -= '0';
  }
  for (int i = 1; i <= n; i++) {
    if (s1[i] == 0 or s2[i] == 0) {
      if (cnt0) ans++;
      cnt0 = 1;
    }
    if (s1[i] == 1 or s2[i] == 1) {
      cnt1 = 1;
    }
    if (cnt1 and cnt0) ans += 2, cnt1 = cnt0 = 0;
  }
  if (cnt0) ans++;
  cout << ans << "\n";
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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>
long long q_pow(long long x, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
using namespace std;
const long long maxn = 1e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
char s1[maxn];
char s2[maxn];
void solve() {
  int n;
  cin >> n;
  cin >> (s1 + 1);
  cin >> (s2 + 1);
  int ans = 0;
  int cnt0 = 0, cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    s1[i] -= '0';
    s2[i] -= '0';
  }
  for (int i = 1; i <= n; i++) {
    if (s1[i] == 0 or s2[i] == 0) {
      if (cnt0) ans++;
      cnt0 = 1;
    }
    if (s1[i] == 1 or s2[i] == 1) {
      cnt1 = 1;
    }
    if (cnt1 and cnt0) ans += 2, cnt1 = cnt0 = 0;
  }
  if (cnt0) ans++;
  cout << ans << "\n";
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```