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;
#include<unistd.h>
#define ll long long
#define pb push_back
#define mp make_pair
#define M 1000000007
#define endl "\n"
bool solve(string str)
{


}
int main() {
	int t;
	cin>>t;
	while(t--)
    {
        int n;
        cin>>n;
        string str1,str2;
        cin>>str1>>str2;
        int one[n],zero[n];
        for(int i=0;i<n;i++)
        {
            one[i]=0;
            zero[i]=0;
        }
       // cout<<str1[0]<<endl;
        if(str1[0]=='0')
            zero[0]=1;
        else one[0]=1;

        for(int i=1;i<n;i++)
        {
            if(str1[i]=='0')

            {   zero[i]=zero[i-1]+1;
                one[i]=one[i-1];
                //cout<<str1[i]<<endl;
            }
            else
            {
                zero[i]=zero[i-1];
                one[i]=one[i-1]+1;
            }

        }

        int flg=0,mainflg=0;
        for(int i=n-1;i>=0;i--)
        {
            if(flg%2==0){
                if(str1[i]!=str2[i])
                {
                    if(zero[i]!=one[i])
                    {
                        mainflg=1;
                        break;
                    }
                    else
                    {
                        flg++;
                    }
                }
            }
            else
            {
                if(str1[i]==str2[i])
                {
                    if(zero[i]!=one[i])
                    {
                        mainflg=1;
                        break;
                    }
                    else
                    {
                        flg++;
                    }
                }
            }
        }
        if(mainflg)
            cout<<"NO\n";
        else cout<<"YES\n";
    }
}





