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;
#define ll long long int
#define pi acos(-1)
#define FAST() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

int main()
{
    FAST();
    ll t;
    cin>>t;
    while(t--)
    {
        string s,ss;
        ll n;
        cin>>n;
        cin>>s>>ss;
        vector<pair<ll,ll>>v;
       ll z=0;
        ll o=0;
        for(ll i=0;i<n;i++)
        {
            if(s[i]=='0')z++;
            else o++;
            v.push_back({z,o});

        }
        ll c=0;
        ll flag=0;
        for(ll i=n-1;i>=0;i--)
        {
            if(c%2==0)
            {
                if(s[i]==ss[i])
                {
                    continue;
                }
                else
                {
                 if(v[i].first!=v[i].second)
                 {
                     flag=1;
                     break;
                 }
                 else
                    c++;
                }
            }
            else
            {
                if(s[i]!=ss[i])
                {
                    continue;
                }
                else
                {
                    if(v[i].first!=v[i].second)
                    {
                        flag=1;
                        break;
                    }
                    else
                    {
                        c++;
                    }
                }
            }
        }
        if(flag==1)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    }

    return 0;
}


