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.
'''def ispalendrome(s):
    i=0
    j=len(s)-1
    while(i<=j):
        if(s[i]!=s[j]):
            return False
        i+=1
        j-=1
    return True
t=int(input())
for _ in range(t):
    s=input()
    n=len(s)
    if(s.count('a')==len(s)):
        print("NO")
    else:
        i=0
        j=len(s)-1
        while(s[j]=='a'):
            j-=1
            i+=1
        
        s=s[:i]+'a'+s[i:]
        print("YES")
        print(s)
        
        while(i<len(s)):
            s=s[0:i]+'a'+s[i:]
           # print('change',s)
            if(ispalendrome(s)):
                s=s[0:i]+s[i+1:]
               # print('orf',s)
                i+=1
                
            else:
                print("YES")
                print(s)
                break
            
    '''
def find(s,j):
    a=0
    b=0
    while(j>=0):
        if(s[j]=='1'):
            a+=1
        else:
            b+=1
        j-=1
    return a,b
'''def convert(s,j):
    while(j>=0):
        if(s[j]=='0'):
            s=s[:j]+'1'+s[j+1:]
        else:
            s=s[:j]+'0'+s[j+1:]
        j-=1
    return s

'''    
t=int(input())
for _ in range(t):
    n=int(input())
    org=input()
    conv=input()
    if(org==conv):
        print("YES")
    else:
        j=n-1
        flip=False
        a,b=find(org,j)
        #print(a,b)
        flag=False
        if(False):
            pass
        else:
            while(j>=0):
                if(not flip):
                    if(conv[j]==org[j]):
                        
                        if(org[j]=='1'):
                            a-=1
                        else:
                            b-=1
                        j-=1
                        continue
                    else:
                        if(a!=b):
                            print("NO")
                           # print(a,b,j)
                            flag=True
                            break
                        else:
                           # print("Fliped")
                            flip=True
                            
                            continue
                else:
                    if(conv[j]!=org[j]):
                        if(org[j]=='1'):
                            a-=1
                        else:
                            b-=1
                        j-=1
                        continue
                    else:
                        if(a!=b):
                            print("NO")
                            flag=True
                           # print(a,b,j)
                            break
                        else:
                            #print("UNfliped")
                            flip=False
                            continue
            if(not flag):
                print("YES")
                         



    
  






    

