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.
t = int(input())
    
    
def solve(n, a, b):
    if (n % 2 == 1):
        if a[-1] != b[-1]:
            return False
        # for i in range(n):
        #     if a[i] != b[i]:
        #         return False
    
        # return True
        n -= 1
    a = [c for c in a]
    c_a = [0]*n
    
    num1 = 0
    for i in range(n):
        if a[i] == 1:
            num1 += 1
    
        c_a[i] = num1
    # print("count", c_a)
    b = [c for c in b]
    flip = False
    for i in range(n-1, -1, -2):
        # print(i, a, flip)
        # print("c",i, c_a[i])
        if flip:
            a[i] = a[i]^1
            a[i-1] = a[i-1]^1
        # print(a[i], b[i], b[i]^1, a[i] == b[i]^1)
        # print(a[i-1], b[i-1], b[i-1]^1, a[i-1] == b[i-1]^1)
        # print("c", c_a[i], c_a[i] == i + 1 - c_a[i])
        if a[i] == b[i]^1 and a[i-1] == b[i-1]^1 and c_a[i] == i + 1 - c_a[i]:
            # print("in if")
            flip = not flip
        elif a[i] == b[i] and a[i-1] == b[i-1]:
            continue
        else:
            return False
    return True
    
    
for _ in range(t):
    n = int(input())
    a = [int(e) for e in input()]
    b = [int(e) for e in input()]
    if solve(n, a, b):
        print("YES")
    else:
        print("NO")