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 os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def lcm(a, b):
    return (a * b) / gcd(a, b)


mod = int(1e9) + 7


def power(k, n):
    if n == 0:
        return 1
    if n % 2:
        return (power(k, n - 1) * k) % mod
    t = power(k, n // 2)
    return (t * t) % mod


def totalPrimeFactors(n):
    count = 0
    if (n % 2) == 0:
        count += 1
        while (n % 2) == 0:
            n //= 2

    i = 3
    while i * i <= n:
        if (n % i) == 0:
            count += 1
            while (n % i) == 0:
                n //= i
        i += 2
    if n > 2:
        count += 1
    return count


# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
#     spf[1] = 1
#     for i in range(2, MAXN):
#         spf[i] = i
#     for i in range(4, MAXN, 2):
#         spf[i] = 2
#
#     for i in range(3, mt.ceil(mt.sqrt(MAXN))):
#         if (spf[i] == i):
#             for j in range(i * i, MAXN, i):
#                 if (spf[j] == j):
#                     spf[j] = i
#
#
# def getFactorization(x):
#     ret = 0
#     while (x != 1):
#         k = spf[x]
#         ret += 1
#         # ret.add(spf[x])
#         while x % k == 0:
#             x //= k
#
#     return ret


# Driver code

# precalculating Smallest Prime Factor
# sieve()

def main():
    for _ in range(int(input())):
        n=int(input())
        f = input()
        s = input()
        ans = 0
        pre = [0, 0]

        for i in range(n):
            #print(s[i], f[i], pre)
            if ord('0') <= ord(s[i]) <= ord('1'):
                if s[i] != f[i]:
                    ans += 2
                    for j in range(2):
                        if pre[j] == 0:
                            ans += j
                            break
                    pre = [0, 0]
                else:
                    if s[i] == '0':
                        if pre[0]:
                            ans += 1
                        else:
                            pre[0] = 1
                            if pre[0]==pre[1]:
                                ans+=2
                                pre=[0, 0]
                    else:
                        if pre[0]:
                            ans += 2
                            pre = [0, 0]
                        else:
                            pre[1] = 1
        for j in range(2):
            if pre[j] == 0:
                ans += j
                break
        print(ans)

    return


if __name__ == "__main__":
    main()
