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.
rr = input
rri = lambda: int(rr())
rrm = lambda: list(map(int, rr().split()))
T = rri()

def solve():
    n = rri()
    a = input()
    b = input()
    dpequal = []
    if a == b:
        return "YES"
    if n % 2 == 1:
        if a[-1] != b[-1]:
            return 'NO'
        else:
            n -= 1

    if a[0] == '1':
        dp = [1]
    else:
        dp = [-1]
    for i in range(0, n, 2):
        aa = a[i:i+2]
        bb = b[i:i+2]
        aam = max(aa.count('0'), aa.count('1'))
        bbm = max(bb.count('0'), bb.count('1'))
        dpequal.append(int(aa==bb))
        if aam != bbm:
            #print(i, aa, bb, n, aa0, bb0)
            return 'NO'


    for i in range(1,n):
        if a[i] == '1':
            temp = 1
        else:
            temp = -1
        dp.append(dp[-1] + temp)
    dpsum = []
    for i in range(1, n, 2):
        dpsum.append(int(dp[i] == 0))
    #print(dp)
    #print(dpequal)
    #print(dpsum)
    if dpsum.count(1) == 0:
        #print('count')
        return 'NO'
    tempi = ''.join(map(str, dpsum)).rindex('1')
    for i in range(tempi+1, len(dpequal)):
        if dpequal[i] != 1:
            #print('last')
            return 'NO'
    i = 0
    while i <= tempi:
        if dpsum[i] == 0:
            #print('in', i)
            start = i
            while dpsum[i] != 1:
                #print('i', i)
                i+=1
                continue
            temp = dpequal[start:i+1]
            #print('temp', temp)
            if temp.count(temp[0]) != len(temp):
                #print('lendiff')
                return 'NO'
            i+=1
        else:
            i+=1

    return 'YES'

for t in range(T):

    print(solve())