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
import math
from collections import defaultdict,Counter,deque

# input=sys.stdin.readline
# def print(x):
#     sys.stdout.write(str(x)+"\n")

import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')

# mod=pow(10,9)+7
t=int(input())
for i in range(t):
	# n,q=map(int,input().split())
	# a=list(map(int,input().split()))
	n=int(input())
	a=input()
	b=input()
	pre=[0]*n
	c=0
	if a[0]=='0':
		c-=1
	else:
		c+=1
	pre[0]=c
	if a[0]==b[0]:
		flag=0
	else:
		flag=1

	for j in range(1,n):
		if a[j]=='0':
			c-=1
		else:
			c+=1
		pre[j]=c

		if a[j]==b[j]:
			if flag:
				if pre[j-1]!=0:
					print("NO")
					break
				flag=0
		else:
			if flag==0:
				if pre[j-1]!=0:
					print("NO")
					break
				flag=1
	else:
		if flag and pre[-1]!=0:
			print("NO")
		else:
			print("YES")
