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 func(n,a,b):
    aone = a.count('1')
    bone = b.count('1')

    if aone != bone:
        print("NO"); return ""
    
    arr = [-1 for x in range(n)]
    one = 0; zero = 0
    for i in range(n):
        if a[i] == '0':
            zero+=1
        elif a[i] == '1':
            one+=1
        
        arr[i] = 1 if zero == one else 0
    flag = 0
    for i in range(n-1,-1,-1):

        if flag == 0:
            if a[i] != b[i] and arr[i] != 1:
                print("NO"); return ""
            elif a[i] == b[i]:
                pass
            elif a[i]!=b[i] and arr[i]==1:
                flag = 1

        elif flag == 1:
            if a[i] == b[i] and arr[i] != 1:
                print("NO"); return ""
            elif a[i] != b[i]:
                pass
            elif a[i]==b[i] and arr[i]==1:
                flag = 0
            
        

    print("YES"); return ""                
        

for _ in range(int(input())):
    n = int(input())
    a = input()
    b = input()

    func(n,a,b)