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.
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[2][(int)3e5 + 10];
//对于单数的情况→判定最后一个是否相等→相等归为双数判断,不相等输出NO
//双数的情况→两个数两个数的判断→0或1的数目必须是偶数→只要有一个不是就输出no
//除此之外,需要检测可以翻转的位置,
//case 1:
// 0 1 1 1 0    在任意两个可以转的区间只要需求一致就可以通过
// 1 0 0 1 0
//case 5:
// 1 0 1
// 0 0 1
//条件1:位数对应上
//遍历,查找每两位是否对应上,标记是否可以翻转,标记是否需要翻转
//实际上遍历的过程中,但凡出现翻转需求突然不一样而上一位没有可翻转标记的,都可以直接停止


int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        scanf("%s%s", s[0], s[1]);
        if (n % 2 == 1 && s[0][n - 1] != s[1][n - 1]) {
            printf("NO\n");
        }
        else {
        	int i = 0;
            int ln = 1, lc = 1, num_1 = 0;
            for (; i < n; i += 2) {
            	//printf("%d : %d %d\n", i, ln, lc);
                int num1 = 0, nown;
                if (s[0][i] == '1') num1++, num_1++;
                if (s[0][i + 1] == '1') num1++, num_1++;
                if (s[1][i] == '1') num1++;
                if (s[1][i + 1] == '1') num1++;
                if (num1 % 2 != 0) {
					printf("NO\n");                	
					break;
                }
                //判断需求
                if (s[0][i] == s[1][i]) nown = 0;
                else nown = 1;
                if (nown != ln && lc != 1) {
					printf("NO\n");                	
					break;
                }
                ln = nown;
                //判断变换
                if (num_1 == i + 2 >> 1) lc = 1;
                else lc = 0;
            }
            if (i == n + 1 || i == n) 
                if (ln == 0 || (ln == 1 && lc == 1))
                    printf("YES\n");
                else printf("NO\n");

        }
    }
    return 0;
}