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<cstdio>
#include<cstdlib>
#include<cstring>
#define NUM 300010
using namespace std;
int n;
int arr[NUM],xe[NUM];
int read() {
    char p = getchar();
    while (!(p == '0' || p == '1')) 
        p = getchar();
    return p - '0';
}
void sovle() {
    int s1 = 0,s0=0;
    int x1=0,x0=0,poi=0;
    for (int i=1;i<=n;i++){
        if (arr[i] == 1 ) s1++;
        if (arr[i] == 0) s0++;
        if (arr[i] != xe[i]) {
            poi = i;
            x1 = s1;
            x0 = s0;
        }
    }
    if (poi == 0){
        puts("YES");
        return ;
    } 
    bool f = false;
    for (int i=poi;i>=1;i--){
        int x = arr[i];
        if (f) x = 1-x;
        if (x != xe[i]){
            if (x1 == x0) f = !f;
            else {
                puts("NO");
                return ;
            }
        }
        x = arr[i];
        if (f) x = 1-x;
        if (x == 1)x1--;
        else x0--;
    }
    puts("YES");
    return ;
}
int main(){
    int t;
    scanf("%d",&t);
    while (t--){
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            arr[i] = read();
        for (int i=1;i<=n;i++)
            xe[i] = read();
        sovle();
    }
    return 0;
}