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 isFlipOrSame(s1, s2):
    num_same = 0
    for i in range(len(s1)):
        if s1[i] == s2[i]:
            num_same += 1
    if num_same == 0 or num_same == len(s1):
        return True
    else:
        return False
    
def isSame(s1, s2):
    num_same = 0
    for i in range(len(s1)):
        if s1[i] == s2[i]:
            num_same += 1
    if num_same == len(s1):
        return True
    else:
        return False

t = int(input().strip())

for i in range(t):
    length = int(input().strip())
    s1 = input().strip()
    s2 = input().strip()
    
    if length <= 1:
        if s1 == s2:
            print("YES")
        else:
            print("NO")
        continue

    num0_1 = [0] * length
    num1_1 = [0] * length
    num0_2 = [0] * length
    num1_2 = [0] * length
    
    if s1[0] == "0":
        num0_1[0] = 1
    else:
        num1_1[0] = 1
        
    if s2[0] == "0":
        num0_2[0] = 1
    else:
        num1_2[0] = 1
    
    for j in range(1, length):
        if s1[j] == "0":
            num0_1[j] = num0_1[j-1] + 1
            num1_1[j] = num1_1[j-1]
        else:
            num0_1[j] = num0_1[j-1]
            num1_1[j] = num1_1[j-1] + 1
            
        if s2[j] == "0":
            num0_2[j] = num0_2[j-1] + 1
            num1_2[j] = num1_2[j-1]
        else:
            num0_2[j] = num0_2[j-1]
            num1_2[j] = num1_2[j-1] + 1
            
    pos = []
    pre_pos = -1        
    for j in range(length):        
        if num0_1[j] == num0_2[j] and num1_1[j] == num1_2[j] and num0_1[j] == num1_1[j]:
            if pre_pos < j-1:
                pos.append(j)
                pre_pos = j
                
    if len(pos) == 0:
        if isSame(s1,s2):
            print("YES")
        else:
            print("NO")
    elif len(pos) == 1:
        if not isFlipOrSame(s1[:pos[0]+1], s2[:pos[0]+1]):
            print("NO")
        elif not isSame(s1[pos[0]+1:], s2[pos[0]+1:]):
            print("NO")
        else:
            print("YES")
    else:
        if not isFlipOrSame(s1[:pos[0]+1], s2[:pos[0]+1]):
            print("NO")
            continue;
        
        check = True
        for p in range(1,len(pos)):
            if not isFlipOrSame(s1[pos[p-1]+1:pos[p]+1], s2[pos[p-1]+1:pos[p]+1]):
                check = False
                break;
        if not isSame(s1[pos[len(pos)-1]+1:], s2[pos[len(pos)-1]+1:]):
            check = False
            
        if check:
            print("YES")
        else:
            print("NO")    