There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.

For example, suppose a=0111010000. 

  * In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00. 
  * In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100. 
  * It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1. 



Can you transform the string a into the string b using some finite number of operations (possibly, none)?

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.

The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.

The sum of n across all test cases does not exceed 3⋅ 10^5.

Output

For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).

Example

Input


5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100


Output


YES
YES
NO
YES
NO

Note

The first test case is shown in the statement.

In the second test case, we transform a into b by using zero operations.

In the third test case, there is no legal operation, so it is impossible to transform a into b.

In the fourth test case, here is one such transformation: 

  * Select the length 2 prefix to get 100101010101. 
  * Select the length 12 prefix to get 011010101010. 
  * Select the length 8 prefix to get 100101011010. 
  * Select the length 4 prefix to get 011001011010. 
  * Select the length 6 prefix to get 100110011010. 



In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
#THIS CODE IS MADE BY "harshest2020"
from sys import *
ws=lambda:map(int,stdin.readline().strip().split())
li=lambda:list(map(int,stdin.readline().strip().split()))
mod=1000000007
def ncr(n, r, p): 
    num = den = 1 
    for i in range(r): 
        num = (num * (n - i)) % p 
        den = (den * (i + 1)) % p 
    return (num * pow(den, p - 2, p)) % p
 
def gcd(a,b):
    if (b == 0): 
         return a 
    return gcd(b, a%b)
def prod(l):
    ans=1
    for i in range(len(l)):
        ans=ans*l[i]
    return ans
def sortindex(l,a):
    c=[]
    if(a==-1):
        rev=True
    else:
        rev=False
    for i in range(len(l)):
        c.append([l[i],i])
    x=sorted(c,reverse=rev)
    print(x)
    c=[]
    for i in range(len(l)):
        c.append(x[i][1])
    return c
def ispalindrome(s):
    c=s[::-1]
    if(c==s):
        return True
    else:
        return False
#CODE STARTS HERE
for _ in range(int(input())):
    size=int(input())
    a=stdin.readline().strip()
    b=stdin.readline().strip()
    dic={0:1}
    t1=0
    t2=0
    for i in range(len(a)):
        if(a[i]=='0'):
            t1+=1
        else:
            t2+=1
        if(t1==t2):
            dic[i+1]=1
    i=0
    flag=0
    ini=0
    
    while(i<size):
        
        if(a[i]==b[i]):
            
            if(flag):
                if(dic.get(ini,0)==1):
                    flag=0
                    if(dic.get(i,0)==1):
                        
                        flag=0
                    else:
                        flag=-1
                        break
                else:
                    flag=-1
                    break
        else:
            if(flag==0):
                ini=i
                if(dic.get(ini,0)==0):
                    flag=-1
                    break
                flag=1
        if(i==(size-1)):
            if(flag):
                if(dic.get(ini,0)==1):
                    flag=0
                    if(dic.get(i+1,0)==1):
                        flag=0
                    else:
                        flag=-1
                        break
                else:
                    flag=-1
                    break
        i+=1
    if(flag==-1):
        print("NO")
    else:
        print("YES")





