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.
import sys
input = sys.stdin.readline
# def mp():return map(int,input().split())
# def lmp():return list(map(int,input().split()))
# def mps(A):return [tuple(map(int, input().split())) for _ in range(A)]
import math
import bisect
from copy import deepcopy as dc
from itertools import accumulate
from collections import Counter, defaultdict, deque
def ceil(U,V):return (U+V-1)//V
def modf1(N,MOD):return (N-1)%MOD+1
inf = int(1e20)
mod = int(1e9+7)
def rle(lst):
    ans = []
    cnt = 1
    ini = lst[0]
    for i in range(1, len(lst)):
        if ini == lst[i]:cnt += 1
        else:
            ans.append((ini, cnt))
            cnt = 1
            ini = lst[i]
    ans.append((ini, cnt))
    return ans

t = int(input())
for _ in range(t):
    n = int(input())
    a = input()[:-1]
    b = input()[:-1]
    c = []
    for i in range(n):
        if a[i] == b[i]:c.append("0")
        else:c.append("1")
    r = rle(c)
    if r[-1][0] == "0":
        r.pop()
    f = True
    for i,j in r:
        if j % 2 != 0:
            f = False
            break
    if not f:
        print("NO")
        continue
    rc = [0]
    for i,j in r:
        rc.append(rc[-1]+j)
    rc = rc[1:]
    a0c = [0]
    a1c = [0]
    for i in range(n):
        if a[i] == "0":
            a0c.append(a0c[-1]+1)
            a1c.append(a1c[-1])
        else:
            a1c.append(a1c[-1]+1)
            a0c.append(a0c[-1])
    for i in rc:
        if a0c[i] != a1c[i]:
            f = False
            break
    if f:print("YES")
    else:print("NO")




