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.
'''
  ____  _                             ____  _             _     
 / ___|(_)_ __ ___  _ __ __ _ _ __   / ___|(_)_ __   __ _| |__  
 \___ \| | '_ ` _ \| '__/ _` | '_ \  \___ \| | '_ \ / _` | '_ \ 
  ___) | | | | | | | | | (_| | | | |  ___) | | | | | (_| | | | |
 |____/|_|_| |_| |_|_|  \__,_|_| |_| |____/|_|_| |_|\__, |_| |_|
                                                    |___/     
  Time: 2021-09-12 20:08:05  
'''
from collections import Counter, deque, defaultdict
from heapq import *
import math 

def inp():
  return int(input())

def mapInput():
  return map(int, input().split())

def arrInput():
  return list(map(int, input().split()))

def task():
  n = inp()
  x = input()
  y = input()
  res, i = 0, 0
  while i < n:
    c = defaultdict(int)
    c[ x[i] ] += 1
    c[ y[i] ] += 1
    if c["0"] > 0 and c["1"] > 0:
      res += 2
      i += 1
      continue 
    inLoop = False 
    while i < n and c["0"] == 0:
      c[ x[i] ] += 1
      c[ y[i] ] += 1
      i += 1
      inLoop = True
    if inLoop:
      i -= 1
    if c["0"] == 0:
      return res
    if c["1"] == 0 and i >= n - 1:
      return res + 1
    if c["1"] == 0:
      if x[i+1] == "1" or y[i+1] == "1":
        res += 2
        i += 2
        continue 
      else:
        res += 1
        i += 1
        continue 
    res += 2
    i += 1
  return res  

for _ in range(inp()):
  print(task())