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 sys
import math
from bisect import bisect_left
import heapq
from collections import deque
from itertools import product, permutations
import random
def II():
	return int(sys.stdin.readline())
 
def LI():
	return list(map(int, sys.stdin.readline().split()))
 
def MI():
	return map(int, sys.stdin.readline().split())
 
def SI():
	return sys.stdin.readline().strip()
 
def C(n, k, mod):
    return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
 
def lcm(a,b):
    return abs(a*b) // math.gcd(a, b)

for _ in range(II()):
	n = II()
	s = list(SI())
	s1 = list(SI())
	ans = 0
	for i in range(len(s)):
		if s[i] == s1[i] == '1':
			if i!=0 and s[i-1] == s1[i-1] == '0':
				s[i-1] = s1[i-1] = ''
				s[i] = '0'
			elif i!=len(s)-1 and s[i+1] == s1[i+1] == '0':
				s[i] = s1[i] = ''
				s[i+1] = '1'
	for i in range(len(s)):
		if s[i] != s1[i]:
			ans+=2
		elif s[i] == '0':
			ans+=1
	print(ans)
