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

zz=1
 
sys.setrecursionlimit(10**5)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]

def fori(n):
	return [fi() for i in range(n)]	
def inc(d,c,x=1):
	d[c]=d[c]+x if c in d else x
def ii():
	return input().rstrip()	
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def dadd(d,p,val):
	if p in d:
		d[p].append(val)
	else:
		d[p]=[val]		
def gi():	
	return [xx for xx in input().split()]
def gtc(tc,*ans):
	print("Case #"+str(tc)+":",*ans)	
def cil(n,m):
	return n//m+int(n%m>0)	
def fi():
	return int(input())
def pro(a): 
	return reduce(lambda a,b:a*b,a)		
def swap(a,i,j): 
	a[i],a[j]=a[j],a[i]	
def bits(i,n):
	p=bin(i)[2:]
	return (n-len(p))*"0"+p	
def prec(a,pre):
	for i in a:
		pre.append(pre[-1]+i)
	pre.pop(0)	
def YN(flag):
	print("YES" if flag else "NO")	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())			
def gh():
	sys.stdout.flush()
def isvalid(i,j,n,m):
	return 0<=i<n and 0<=j<m 
def bo(i):
	return ord(i)-ord('a')	
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)



t=fi()
INF=10**18
uu=t
mod=10**9+7

			
while t>0:
	t-=1
	n=fi()
	a=[ii(),ii()]
	dp=[[0]*3 for i in range(n)]
	for i in range(n):
		p=set([int(a[0][i]),int(a[1][i])])
		p=list(p)
		if len(p)==2:
			dp[i][2]=max(dp[i][2],max(dp[i-1])+2)
		elif p[0]==0:
			if i==0:
				dp[i][1]=1
				continue
			dp[i][1]=max(dp[i][1],max(dp[i-1])+1)
			dp[i][2]=max(dp[i-1][0]+2,max(dp[i-1][2],dp[i-1][0])+1)
		else:
			if i==0:
				dp[i][0]=0
				continue
			dp[i][0]=max(dp[i-1])
			dp[i][2]=max([dp[i-1][2],dp[i-1][0],dp[i-1][1]+1])
	print(max(dp[n-1]))		