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.
# link: https://codeforces.com/contest/1566/problem/C

for _ in range(int(input())):
    n = int(input())
    a = list(input())
    b = list(input())
    if len(set(a + b)) == 1:
        if "1" in set(a + b):
            print(0)
        else:
            print(n) 
    else:
        ans = 0
        diff = set()
        same = set()
        i = 0
        for first, second in zip(a, b):
            if first != second:
                diff.add(i)
            else:
                same.add(i)    
            i += 1
        i = 0    
        for f, s in zip(a, b):    
            if f == s:
                if i in same and (i-1) in same and a[i-1] != a[i]:
                    ans += 2
                    same.remove(i-1)
                    same.remove(i)
                elif i in same and (i+1) in same and a[i+1] != a[i]:        
                    ans += 2
                    same.remove(i+1)
                    same.remove(i)
                elif i in same and a[i] == '1':
                    if (i-1) in diff:
                        diff.remove(i-1)
                        same.remove(i)
                        ans += 2
                    elif (i+1) in diff:
                        diff.remove(i+1)
                        same.remove(i)
                        ans += 2                 
            i += 1        
        ans += 2*len(diff)
        for i in same:
            if a[i] == "0":
                ans += 1
        print(ans)         