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 = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
# from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# from itertools import groupby
# sys.setrecursionlimit(10000+10) #this is must for dfs
# ______________________________________________________________________________________________________
# segment tree for range minimum query
# n = int(input())	
# a = list(map(int,input().split()))
# st = [float('inf') for i in range(4*len(a))]
# def build(a,ind,start,end):
# 	if start == end:
# 		st[ind] = a[start]
# 	else:
# 		mid = (start+end)//2
# 		build(a,2*ind+1,start,mid)
# 		build(a,2*ind+2,mid+1,end)
# 		st[ind] = min(st[2*ind+1],st[2*ind+2])
# build(a,0,0,n-1)
# def query(ind,l,r,start,end):
# 	if start>r or end<l:
# 		return float('inf')
# 	if l<=start<=end<=r:
# 		return st[ind]
# 	mid = (start+end)//2
# 	return min(query(2*ind+1,l,r,start,mid),query(2*ind+2,l,r,mid+1,end))
# ______________________________________________________________________________________________________
# Checking prime in O(root(N))
# def isprime(n):
#     if (n % 2 == 0 and n > 2) or n == 1: return 0
#     else:
#         s = int(n**(0.5)) + 1
#         for i in range(3, s, 2):
#             if n % i == 0:
#                 return 0
#         return 1
# def lcm(a,b):
#   return (a*b)//gcd(a,b)
# ______________________________________________________________________________________________________
# nCr under mod
# def C(n,r,mod = 10**9+7):
# 	if r>n: return 0
# 	if r>n-r: r = n-r
# 	num = den = 1
# 	for i in range(r):
# 		num = (num*(n-i))%mod
# 		den = (den*(i+1))%mod
# 	return (num*pow(den,mod-2,mod))%mod
# def C(n,r):
# 	if r>n:
# 		return 0
# 	if r>n-r:
# 		r = n-r
# 	ans = 1
# 	for i in range(r):
# 		ans = (ans*(n-i))//(i+1)
# 	return ans
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 2*10**5+10
# spf = [i for i in range(M)]
# def spfs(M):
#   for i in range(2,M):
#       if spf[i]==i:
#           for j in range(i*i,M,i):
#               if spf[j]==j:
# 	              spf[j] = i
#   return
# spfs(M)
# ______________________________________________________________________________________________________
# def gtc(p):
#     print('Case #'+str(p)+': ',end='')
# ______________________________________________________________________________________________________
# # not my code
# # To avoid runtime error while using recursive functions
# # Instead of sys.setrecursionlimit(Max_depth) use this
# def bootstrap(f, stack=[]):
#     from types import GeneratorType
#     def wrappedfunc(*args, **kwargs):
#         if stack:
#             return f(*args, **kwargs)
#         else:
#             to = f(*args, **kwargs)
#             while True:
#                 if type(to) is GeneratorType:
#                     stack.append(to)
#                     to = next(to)
#                 else:
#                     if not stack:
#                         break
#                     stack.pop()
#                     to = stack[-1].send(to)
#             return to
#     return wrappedfunc
# #### use this in the previous line of the recursive function '@bootstrap'
# ______________________________________________________________________________________________________
tc = 1
tc = int(input())
for test in range(1,tc+1):
	n = int(input())
	a = str(input())
	b = str(input())
	ans = []
	i  =  0
	f = True
	o,z = 0,0
	f1 = False
	lol = []
	while(i<n):
		cnt = 0
		while(i<n and a[i]!=b[i]):
			if a[i]=='0':
				o+=1
			else:
				z+=1
			i+=1
			cnt+=1
		if cnt>0 and o!=z:
			f = False
			break
		if f1 and lol[-1]:
			f = False
			break
		cnt = 0
		while(i<n and a[i]==b[i]):
			if a[i]=='0':
				o+=1
			else:
				z+=1
			i+=1
			cnt+=1
			f1 = True
		lol.append(o!=z)
		# if cnt>0 and o!=z:
		# 	f = False
		# 	break
	print("YES" if f else "NO")