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;
//#pragma GCC optimize "trapv"
#define ll long long

void solve() {
    int n; string a,b; cin>>n>>a>>b;
    int f0=0,f1=0;
    vector<bool> marked(n);
    for (int i=0;i<n;i++) {
    	if (a[i]=='0')
    		f0++;
    	else
    		f1++;
    	if (f0==f1) {
    		//cout<<i<<endl;
    		bool allsame = true;
    		bool toggled = true;
    		for (int j=i;j>=0 && !marked[j];j--) {
    			if (a[j]!=b[j]) 
    				allsame = false;
    			if (a[j]==b[j])
    				toggled = false;
    		}
    		if (allsame || toggled) 
    			marked[i] = true;
    		else {
    			cout<<"NO\n";
    			return;
    		}
    	}
    }
    bool same = true;	//or marked
    for (int i=n-1;i>=0 && !marked[i];i--) {
    	if (a[i]!=b[i]) {
    		same = false;
    		break;
    	}
    }
    if (same) cout<<"YES\n";
    else cout<<"NO\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t; cin>>t;
    while (t--) 
        solve();
    return 0;
}