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.
#region Header
#!/usr/bin/env python3
# from typing import *

import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq


def input():
    return sys.stdin.readline()[:-1]


# sys.setrecursionlimit(1000000)
#endregion

# _INPUT = """# paste here...
# """
# sys.stdin = io.StringIO(_INPUT)

YES = 'YES'
NO = 'NO'


def solve(N, A, B):

    T_A0, T_A1, T_B0, T_B1 = [], [], [], []
    n_A0, n_A1, n_B0, n_B1 = 0, 0, 0, 0
    for i in range(N):
        if A[i] == '0':
            n_A0 += 1
        else:
            n_A1 += 1
        T_A0.append(n_A0)
        T_A1.append(n_A1)

        if B[i] == '0':
            n_B0 += 1
        else:
            n_B1 += 1
        T_B0.append(n_B0)
        T_B1.append(n_B1)

    flag = True # True: same; False: different
    for i in reversed(range(N)):
        if flag and A[i] != B[i]:
            flag = False
            if T_A0[i] != T_A1[i] or T_B0[i] != T_B1[i]:
                return False
        elif (not flag) and A[i] == B[i]:
            flag = True
            if T_A0[i] != T_A1[i] or T_B0[i] != T_B1[i]:
                return False
    return True

def main():
    T0 = int(input())
    for _ in range(T0):
        N = int(input())
        A = input()
        B = input()
        ans = solve(N, A, B)
        if ans:
            print('YES')
        else:
            print('NO')


if __name__ == '__main__':
    main()
