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.
#!/usr/local/bin/python3

import bisect
import collections
import functools
import heapq
import itertools
import math
import random
import sys
import time
input = sys.stdin.readline
# sys.setrecursionlimit(500005)
def ri(): return int(input())
def rf(): return list(map(float, input().split()))
def rl(): return list(map(int, input().split()))
def rs(): return input().rstrip()
def out_list(x): return ' '.join(map(str, x))

MOD = 10**9 + 7
INF = float('inf')

def solve_case():
    n = ri()
    A = [rs() for _ in range(2)]
    res = 0
    for rep in range(2):
        cur = 0
        i = 0
        while i < n:
            if int(A[0][i]) + int(A[1][i]) == 1:
                cur += 2
                i += 1
            elif int(A[0][i]) + int(A[1][i]) == 2:
                if i + 1 < n and int(A[0][i + 1]) + int(A[1][i + 1]) == 0:
                    cur += 2
                    i += 2
                else:
                    i += 1
            elif int(A[0][i]) + int(A[1][i]) == 0:
                if i + 1 < n and int(A[0][i + 1]) + int(A[1][i + 1]) == 2:
                    cur += 2
                    i += 2
                else:
                    cur += 1
                    i += 1
        res = max(res, cur)
        for i in range(2):
            A[i] = A[i][::-1]
    print(res)

if __name__ == '__main__':
    # start_time = time.time()
    # T = 1
    T = int(input())
    for test_case in range(1, T + 1):
        solve_case()
    # end_time = time.time()
    # print('Runtime (s): {}'.format(end_time - start_time))
