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,math,bisect
# sys.setrecursionlimit(10**6)
from random import randint
inf = float('inf')
mod = 10**9+7
"========================================"
def nCr(n, r):
    return (fact(n) / (fact(r)
                * fact(n - r)))
def fact(n):
    res = 1
    for i in range(2, n+1):
        res = res * i
    return res
def lcm(a,b):
    return int((a/math.gcd(a,b))*b)
def gcd(a,b):
    return int(math.gcd(a,b))
def tobinary(n):
    return bin(n)[2:]
def binarySearch(a,x):
    i = bisect.bisect_left(a,x)
    if i!=len(a) and a[i]==x:
        return i
    else:
        return -1
def lowerBound(a, x):
    i = bisect.bisect_left(a, x)
    if i:
        return (i-1)
    else:
        return -1
def upperBound(a,x):
    i = bisect.bisect_right(a,x)
    if i!= len(a)+1 and a[i-1]==x:
        return (i-1)
    else:
        return -1
def primesInRange(n):
    ans = []
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    for p in range(2, n+1):
        if prime[p]:
            ans.append(p)
    return ans
def primeFactors(n):
    factors = []
    while n % 2 == 0:
        factors.append(2)
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2):
        while n % i== 0:
            factors.append(i)
            n = n // i
    if n > 2:
        factors.append(n)
    return factors
def isPrime(n,k=5):
    if (n <2):
        return True
    for i in range(0,k):
        a = randint(1,n-1)
        if(pow(a,n-1,n)!=1):
            return False
    return True
"========================================="
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
from collections import deque,defaultdict,Counter
from heapq import heappush, heappop,heapify
import string


for _ in range(int(input())):
    n= int(input())
    arr=input()
    brr=input()
    cnt = 0
    visited = [0]*n
    for i in range(n):
        if visited[i]:
            continue
        curr = [arr[i],brr[i]]
        if '0' in curr and '1' in curr:
            cnt+=2
            visited[i]=1
        else:
            if i!=n-1:
                next = [arr[i+1],brr[i+1]]
                if '0' in next and '1' in next:
                    pass
                elif ('0' in curr and '1' in next) or ('1' in curr and '0' in next):
                    cnt+=2
                    visited[i]=1
                    visited[i+1]=1
                else:
                    if '0' in curr:
                        cnt+=1
                    visited[i]=1
            else:
                if '0' in curr:
                    cnt+=1
    print(cnt)
