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<vector>
using namespace std;
int t;
int main(){
    cin>>t;
    while(t--){
        int n;cin>>n;
        string a,b;
        vector<int> v;
        v.push_back(-1);
        cin>>a>>b;
        int a1=0,b1=0,a0=0;
        for (int i=0;i<n;i++){
            if (a[i]=='1') a1++;
            else a0++;
            if (b[i]=='1') b1++;
            if (a1==a0) v.push_back(i);
        }
        if (a1!=b1){
            cout<<"NO"<<endl;
            continue;
        }
        if (a==b){
            cout<<"YES"<<endl;
            continue;
        }
        int vlen=v.size();
        bool ok=true;
        //for(int u:v)cout<<u;
        for (int i=0;i<vlen-1;i++){

            int now=v[i]+1;
            int nxt=v[i+1];
            bool same;
            if (a[now]==b[now]) same=true;
            else same=false;
            //(same)?cout<<"true":cout<<"false";
            for (int j=now;j<=nxt;j++){
                if (a[j]==b[j] && same==false){ ok=false; break;}
                if (a[j]!=b[j] && same==true){ ok=false; break;}
            }
        }
        for (int i=v[vlen-1]+1;i<n;i++){
            if (a[i]!=b[i]) ok=false;
        }
        if (v.size()==1 && a!=b) cout<<"NO"<<endl;
        else if (ok) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;

    }
}