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.
for i in range(int(input())):
    n_cols = int(input())
    r1, r2 = input(),input()
    temp,temp0 = False,False
    su = 0
    for j in range(n_cols):
        row = r1[j] + r2[j]
        if row == '11':
            if temp0: # 00 11
                su+=2
                temp0 = False
                temp = False
            else: ## wasnt00 now 11 - keep for next
                j+=1
                temp = True
        if row != '11':
            if temp: # was11 now:xx
                su += 2
                temp = False
            else:
                if row == '00': #wasnt11 now00
                    if temp0: # was00 now00
                        su += 1 ##
                        if j+1 == n_cols: su+=1 #last col - add for prev
                        # elif r1[j+1]+r2[j+1] == '00': su+=1 # 2 00 in a row
                    elif j+1 != n_cols and (r1[j+1] == '1' and r2[j+1] == '1'): #not last and 11 next
                        temp0 = True #00 but skip for now
                        j+=1
                    else:
                        su+=1 # 00 and not last and not prev 00
                else: su += 2 #wasnt 11 now 01/10
    print(su)