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.

from __future__ import division
import sys
input = sys.stdin.readline
import math
from math import sqrt, floor, ceil
from collections import Counter, deque, defaultdict
from copy import deepcopy as dc


############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def insr2():
    s = input()
    return(s.split(" "))

def prime_factorization(n):

    if n == 1:
        return [1]

    ans=[]
    i = 2
    cap = sqrt(n)
    while i <= cap:
        if n % i == 0:
            ans.append(i)
            n = n//i
            cap=sqrt(n)
        else:
            i += 1
    if n > 1:
        ans.append(n)
    return ans

def binomial(n, k):
    if n == 1 or n == k:
        return 1

    if k > n:
        return 0       
    else:
        a = math.factorial(n)
        b = math.factorial(k)
        c = math.factorial(n-k)
        div = a // (b * c)
        return div 

#
# r = raw_input()
        
for _ in range(inp()):
    n = inp()
    f = True
    ans = True
    a = raw_input()
    b = raw_input()

    i = 0
    pre = []
    val = []

    ind = set()

    if a[0] == b[0]:
        f = True
    else:
        f = False

    if a[0] == '0':
        c0, c1 = 1, 0
    else:
        c0, c1 = 0, 1

    for i in range(1, n):
        if a[i] == '0':
            c0 += 1
        else:
            c1 += 1

        if c0 == c1:
            ind.add(i)

        if a[i] == b[i]:
            if f == True:
                continue
            else:
                if (i-1) in ind:
                    f = True
                    continue
                else:
                    ans = False
                    break     
        elif a[i] != b[i]:
            if f == False:
                continue
            else:
                if (i-1) in ind:
                    f = False
                    continue
                else:
                    ans = False
                    break
        


    if f == False:
        if n-1 not in ind:
            print 'NO'
            continue

    if ans:
        print 'YES'
    else:
        print 'NO'


