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 math

def int_input():
    a = int(input())
    return a

def int_list_input():
    a = list(map(int,input().split()))
    return a

def solve():
    n = int_input()
    a = input()
    b = input()
    ans = 0
    i = 0
    s = ""
    
    while i < n:
        if a[i]!=b[i]:
            i += 1
            ans += 2
            s += '!'
        else:
            s += a[i]
            i += 1
    
    i = 0
    n = len(s)
    #print(s)
    while i < n:
        if s[i] == '!':
            i += 1
            continue
        j = i+1
        while j<n and s[j]==s[i]:
            j += 1
        if j<n and s[j]=='!':
            i = j+1
            continue
        if j!=n:
            ans += 2
        elif s[i]!='!' and s[j-1]!=s[i]:
            ans += 2
        elif s[i] == '0':
            ans += (j-i)
        i = j+1
        #print(ans,i)
    print(ans)
    
    
if __name__ == '__main__':
    
    for _ in range(int(input())):
        solve()
    
    #print("Case",str(n)+":",str(ans))