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;
typedef long long ll;
typedef long double ld;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define mod 1000000007
#define modx 998244353
#define inf 2000000000000000000
#define mem(name,val) memset(name,val,sizeof(name))
#define precise cout<<fixed<<setprecision(15)
const long double pi = acos(-1);
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); 
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        string a,b;
        cin>>a>>b;
        int f=1,fc=0,c1=0,c0=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]=='0')
            c0++;
            else
            c1++;
        }
        for(int i=n-1;i>=0;i--)
        {
            if((a[i]!=b[i]&&fc%2==0)||(a[i]==b[i]&&fc%2==1))
            {
                if(c0==c1)
                {
                    fc++;
                    swap(c0,c1);
                }
                else
                {
                    f=0;
                    break;
                }
            }
            if(fc%2==0)
            {
                if(a[i]=='0')
                c0--;
                else
                c1--;
            }
            else
            {
                if(a[i]=='0')
                c1--;
                else
                c0--;
            }
        } 
        if(f==1)
        cout<<"YES\n";
        else
        cout<<"NO\n";     
    }
    return 0;
}
