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.
import sys
from functools import lru_cache

input = sys.stdin.readline


def cin_int():
    return int(input())


def cin_ls_int():
    return list(map(int, input().split()))


def cin_ls():
    return input().split()


def main():
    v = {(True, True): 2,
         (True, False): 1,
         (False, True): 0,
         (False, False): 0,
         }

    @lru_cache(None)
    def dp(idx, _01):
        if idx == len(s1):
            return v[_01]
        a, b = int(s1[idx]), int(s2[idx])
        l = list(_01)
        l[a] = True
        l[b] = True
        return max(
            dp(idx + 1, tuple(l)),
            dp(idx + 1, (False, False)) + v[tuple(l)]
        )

    def sv():
        idx = 0
        ans = 0
        pre = [False, False]
        while idx != len(s1):
            a, b = int(s1[idx]), int(s2[idx])
            if a + b == 2:
                if pre[0]:
                    pre = [False, False]
                    ans += 2
                else:
                    pre[1] = True
            elif a + b == 1:
                ans += v[tuple(pre)]
                pre = [False, False]
                ans += 2
            else:
                if pre[1]:
                    pre = [False, False]
                    ans += 2
                elif pre[0]:
                    ans += 1
                else:
                    pre[0] = True
            idx += 1
        return ans + v[tuple(pre)]

    for _ in range(cin_int()):
        input()
        s1 = str(input()).strip('\n').strip()
        s2 = str(input()).strip('\n').strip()
        print(sv())


if __name__ == '__main__':
    main()
