#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
const long long p = 998244353;
long long qp(long long x, long long y) {
  long long res = 1;
  for (long long t = x; y; y >>= 1, t = t * t % p)
    if (y & 1) res = res * t % p;
  return res;
}
char s[1000003], t[1000003];
long long a[1000003];
signed main() {
  for (long long T = read(); T--;) {
    long long n = read();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    for (long long i = 1; i <= n; ++i) a[i] = (s[i] == '1') + (t[i] == '1');
    long long cur = 0, qwq = 0, ans = 0;
    for (long long i = 1; i <= n; ++i)
      if (a[i] == 1) {
        ans += cur;
        ans += 2;
        cur = 0;
        qwq = 0;
      } else if (a[i] == 0) {
        ans += cur;
        if (qwq)
          qwq = 0, ans += 2, cur = 0;
        else
          cur = 1;
      } else {
        if (cur == 1)
          ans += 2, cur = 0;
        else
          qwq = 1;
      }
    printf("%lld\n", ans + cur);
  }
  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>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
const long long p = 998244353;
long long qp(long long x, long long y) {
  long long res = 1;
  for (long long t = x; y; y >>= 1, t = t * t % p)
    if (y & 1) res = res * t % p;
  return res;
}
char s[1000003], t[1000003];
long long a[1000003];
signed main() {
  for (long long T = read(); T--;) {
    long long n = read();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    for (long long i = 1; i <= n; ++i) a[i] = (s[i] == '1') + (t[i] == '1');
    long long cur = 0, qwq = 0, ans = 0;
    for (long long i = 1; i <= n; ++i)
      if (a[i] == 1) {
        ans += cur;
        ans += 2;
        cur = 0;
        qwq = 0;
      } else if (a[i] == 0) {
        ans += cur;
        if (qwq)
          qwq = 0, ans += 2, cur = 0;
        else
          cur = 1;
      } else {
        if (cur == 1)
          ans += 2, cur = 0;
        else
          qwq = 1;
      }
    printf("%lld\n", ans + cur);
  }
  return 0;
}

```