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 solver(a, b):
    
    a = list(map(int,list(a)))
    b = list(map(int,list(b)))
    
    swapped_version = [1 if el == 0 else 0 for el in a]

    if a == b:
        return 'YES'
    
    if sum(a) != sum(b):
        return 'NO'

      
    answers = []
    valid_pref = []
    cur_sum = 0
    for i in range(0, len(a)):
        if a[i] == 1:
            cur_sum += 1
        elif a[i] == 0:
            cur_sum -= 1
            
        # print(i, cur_sum, a)
        if cur_sum == 0:
            valid_pref.append(i)
                
    valid_pref =  [len(a)] + valid_pref[::-1] + [0]

    
    swapped_bool = False

    for x,y in zip(valid_pref[1:],valid_pref[:-1]):

        #swapped = [1 if el == 0 else 0 for el in a[:(x+1)]] + a[x+1:]

        if swapped_bool:
            compare = [1 if el == 0 else 0 for el in a[x:y]].copy()
        else:
            compare = a[x:y].copy()
            
            
        no_cond = compare != b[x:y]# and swapped[x:y] != b[x:y]
        no_cond = no_cond and  not (compare[0] != b[x] and compare[1:y] == b[x+1:y])
        
        #print(swapped[x:y] != b[x:y], not (a[x] != b[x] and a[x+1:y] == b[x+1:y]))
        if no_cond :
            return 'NO'
        if compare != b[x:y]:

            swapped_bool = not swapped_bool
            #a = swapped.copy()

    return 'YES'

#solver('000111','110100')
    
#x = list(map(int,list('010101010101')))
#y = list(map(int, list('100110011010')))

#update_check(x, y)

t = int(input())
answers = []

#5
#10
#0111010000
#0100101100
#4
#0000
#0000
#3
#001a
#000
#12
#010101010101
#100110011010
#6
#000111
#110100

ls = []
a_s = []
b_s = []
for i in range(t):
    _ = input()
    a_s.append(input())
    b_s.append(input())
    
    ans = solver(a_s[-1], b_s[-1])
    answers.append(ans)

print('\n'.join(answers))