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 <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int t,n;
string a,b;
void solve(){
        
}
void sieve() {
	int a[1010];
	for (int i=0; i<1e3; i++) cin>>a[i];
	cout<<a[10];
}
signed main(){
    int t;
    cin>>t;
    while (t--) {
    	cin >> n >> a >> b;
        vector <int> vec;
        map<int,int> m;
        vec.push_back(-1);
        int dem[2] = {0};
        int den[2] = {0};
        bool check = 0;
        for(int i = 0; i < n; i++){
            dem[(int)(a[i] - 48)]++;
            if(dem[0]==dem[1]){
                m[i]++;
                vec.push_back(i);
            }
        }
        for(int i = 1; i < vec.size(); i++){
            int cug = 0, ngc = 0;
            for(int j = vec[i - 1] + 1; j <= vec[i]; j++){
                if(a[j]==b[j]) cug++;
                else ngc++;
            }
            if(ngc!=0 && cug!=0){
                check = 1;
                break;
            }
        }
        if(check==1) cout << "NO" << endl;
        else{
            for(int i = vec[vec.size() - 1] + 1; i < n; i++){
                if(a[i]!=b[i]){
                    check = 1;
                    break;
                }
            }
            if(check==1) cout << "NO" << endl;
            else cout << "YES" << endl;
        }
	}
}