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_right as br

def int_arr(): return list(map(int, sys.stdin.readline().split()))

def str_arr(): return list(map(str, sys.stdin.readline().split()))

def input(): return sys.stdin.readline().strip()

for _ in range(int(input())):
	n=int(input())
	a=input()
	b=input()
#	if n==1:
#		if a=='1' and b=='1':
#			print(0)
#		if a=='0' and b=='0':
#			print(1)
#		else:
#			print(2)
#		continue
#	k=n
#	if n%2!=0:
#		k-=1
#	ans=0
#	for i in range(0,k,2):
#	#	print(i)
#		c=0
#		for x in range(i,i+2):
#			if a[x]=="0":
#				c+=1
#			if b[x]==0:
#				c+=1
#		if c==4:
#			ans+=4
#		elif c==3:
#			ans+=3
#		elif c==2 or c==1:
#			ans+=2
#	if n%2!=0:

#		if a[-1]=='0' and b[-1]=='0':
#			ans+=1
#		else:
#			ans+=2
#	print(ans)



#	flag=1 for 00
#	flag=2 for 11
	ans=0
	dict0={}
	dict1={}
	
	for i in range(n):
		if a[i]=='0' and b[i]=='0':
			dict0[i]=True
		elif a[i]=='1' and b[i]=='1':
			dict1[i]=True
		else:
			ans+=2
	
	for i in dict0.keys():
		if dict1.get(i-1)==True:
			ans+=2
		elif dict1.get(i+1)==True:
			dict1[i+1]=0
			ans+=2
		else:
			ans+=1
	print(ans)
	

		
		
		
	