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.
#!/usr/bin/python

import sys
from sys import stdin
#use ./ex1.py < input_ex1.py
    
def is_good(a_part, b_part):
    if a_part == b_part:
        return True
    b_part_copy = ''
    for b in b_part:
        if b == '0':
            b_part_copy += '1'
        else:
            b_part_copy += '0'
    if a_part == b_part_copy:
        return True
    return False
    
def solve(N, A, B):
    a_start = -1
    a_status = 0
    b_status = 0
    for i in range(N):
        if A[i] == '0':
            a_status -= 1
        else:
            a_status += 1
        if B[i] == '0':
            b_status -= 1
        else:
            b_status += 1
        if a_status == 0:
            if b_status == 0:
                a_part = A[(a_start+1):(i+1)]
                b_part = B[(a_start+1):(i+1)]
                if not is_good(a_part, b_part):
                    return "NO"
                a_start = i
            else:
                return "NO"
    if a_start == (N-1):
        return "YES"
    else:        
        a_part = A[(a_start+1):]
        b_part = B[(a_start+1):]
        if a_part == b_part:
            return "YES"
        else:
            return "NO"
    
def run():
    out = ""    
    T = int(input())
    for i in range(T):
        N = int(input())
        A = input()
        B = input()
        print(solve(N, A, B))

run()

