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 sys import stdin
 
 
N = int(stdin.readline())
for t in range(N):
   #print()
   #print(t+1)
   
   n =  int(stdin.readline())
   s1 = stdin.readline().strip()
   s2 = stdin.readline().strip()
   #print(s1)
   #print(s2)
   splitPoints = []
   c = 0
   for i in range(n):
      if i%2==0 and c == i//2:
         splitPoints.append(i)
      if s1[i] == '1': c+=1
   if n%2==0 and c== n//2: splitPoints.append(n)
   
   #print('splits:' + str(splitPoints))
   if len(splitPoints) ==1:
      if s1 !=s2: print('NO')
      else: print('YES')
   else:
      lastPart = s1[splitPoints[-1]:]
      lastPart2 = s2[splitPoints[-1]:]
      #print(lastPart)
      #print(lastPart2)
      bad = False
      if lastPart != lastPart2: 
         print('NO')
      else:
         j = 0
         while j<len(splitPoints)-1:
            act1 = s1[splitPoints[j] : splitPoints[j+1]]
            act1_flip = ''.join(['0' if s=='1' else '1' for s in act1])
            act2 = s2[splitPoints[j] : splitPoints[j+1]]
      
            if act1 != act2 and act1_flip != act2:
               print('NO')
               bad = True
               break
            j+=1
         if bad == False:
            print('YES')
            
                
      
         
          