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.
def chck(a1, a2, b1, idx, a, b, n):
    s1 =[] 
    s2 = []
    s3 = []
    z = o = 0
    ind = 0
    for i in range(n):
        ai = a[i]
        s1.append(ai)
        s2.append(1 - ai)
        s3.append(b[i])
        if ai == 0: z += 1
        else: o += 1
        if z == o:
            a1.append(s1)
            a2.append(s2)
            b1.append(s3)
            idx.append(i)
            s1 =[] 
            s2 = []
            s3 = []
            ind += 1

    return ind

t = int(input())
for i in range(t):
    n = int(input())
    a = [int(x) for x in input()]
    b = [int(x) for x in input()]
    a1 = []
    a2 = []
    b1 = []
    idx = []
    ind = chck(a1, a2, b1, idx, a, b, n)
    m = 0
    p = -1
    for i in range(ind):
        p = idx[i] - p
        for x in range(p): 
            if (a1[i][x] != b1[i][x]):
                m = 1
                break
        if m == 1:
            m = 0
            for x in range(p): 
                if (a2[i][x] != b1[i][x]):
                    m = 1
                    break
        if m == 1: break
        p = idx[i]
    if m == 0:
        for i in range(p + 1, n):
            if (a[i] != b[i]): 
                m = 1
                break

    if m == 0: print("YES")
    else: print("NO")
    
