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 sys import stdin,stdout
def INPUT():return list(int(i) for i in stdin.readline().split())
def inp():return stdin.readline()
def out(x):return stdout.write(x)
from math import *
from decimal import *
INT_MAX=10**13
# from bisect import bisect_left
# import sys
# MOD=10**9+7
# from random import randint as R
# from heapq import heappush,heappop,heapify
#320660CC
mod=998244353
#=============================================
def ceil(a,b):
    return (a+b-1)//b
def pow(a,n):
    res=1
    while(n):
        if n&1:
            res=(res*a)%mod
        a=a*a
        a%=mod
        n>>=1
    return res
for _ in range(int(input())):
    n=int(input())
    a=input()
    b=input()
    n=len(a)
    paired=[False]*(n+1)
    ans=0
    for i in range(n):
        if int(a[i])^int(b[i])==1:
            ans+=2
        elif i==0:
            if a[i]=='0' and b[i]=='0':
                ans+=1
        else:
            if a[i]=='0' and b[i]=='0':
                if a[i-1]=='1' and b[i-1]=='1' and paired[i-1]==False:
                    ans+=2
                    paired[i]=True
                else:
                    ans+=1
            else:
                if a[i-1]=='0' and b[i-1]=='0' and paired[i-1]==False:
                    ans+=1
                    paired[i]=True
    print(ans)
    #print()
