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.
from collections import defaultdict
import math
 
# from functools import reduce
# from math import comb, gcd # n! / (k! * (n - k)!)
# from collections import defaultdict
# from sys import stdout
 
 
NO = 'NO'
YES = 'YES'
MOD = 1000000007
ALPH = 'abcdefghijklmnopqrstuvwxyz'
 
def input_int():
    return int(input())
 
def input_list():
    return input().split(' ')
 
def input_list_int():
    return list(map(int, input_list()))
 
def list_int_to_str(a, sep = ' '):
    return sep.join(str(x) for x in a) 
 
global_ans = []
def pr(a):
    global_ans.append(a)
 
def z_func(s):
    n = len(s)
    z = [0] * n
    l = r = 0
    for i in range(1, n):
        if r >= i:
            z[i] = min(z[i - l], r - i + 1)
        while z[i] + i < n and s[z[i]] == s[z[i] + i]:
            z[i] += 1
        if i + z[i] - 1 > r:
            l = i
            r = i + z[i] - 1
    return z
 
def is_prime(n):
    d = 2
    while n % d != 0:
        d += 1
    return d == n


def solve():
    n = input_int()
    a = [
        input(),
        input()
    ]
    ans = 0
    cnt = {}
    for i in range(n - 1):
        if i not in cnt and a[0][i] == a[1][i] and a[0][i + 1] == a[1][i + 1] and a[0][i] != a[0][i + 1]:
            ans += 2
            cnt[i] = 1
            cnt[i + 1] = 1

    tmp = defaultdict(int)
    for i in range(n):
        if i in cnt:
            tmp = defaultdict(int)
            continue
        tmp[a[0][i]] += 1
        tmp[a[1][i]] += 1
        if tmp['0'] > 0 and tmp['1'] > 0:
            ans += 2
            tmp = defaultdict(int)
        elif tmp['0'] > 0:
            ans += 1
            tmp = defaultdict(int)

    pr(ans)

query_count = 1
query_count = input_int()
 
 
while query_count:
    query_count -= 1
    solve()
 
for global_ans_line in global_ans:
    print(global_ans_line)
 
 