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 fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define lf(i, a, b) for(ll i=a;i<b;i++)
#define lb(i, a, b) for(ll i=a;i>b;i--)
#define pb push_back

void solve(){
    int n, o = 0, z = 0, flag = 0, ans = 1;
    cin>>n;
    string a, b;
    stack <int> s;
    cin>>a>>b;
    lf (i, 0, n){
        if (a[i] == '1')
            o++;
        else    
            z++;
        if (o == z)
            s.push(i);
    }
    lb (i, n, -1){
        if (!flag){
            if (a[i] == b[i]){
                if (!s.empty() && s.top() == i)
                    s.pop();
                continue;
            }
            if (!s.empty() && s.top() == i){
                flag = 1;
                s.pop();
                continue;
            }
            ans = 0;
            break;
        }
        else{
            if (a[i] != b[i]){
                if (!s.empty() && s.top() == i)
                    s.pop();
                continue;
            }
            if (!s.empty() && s.top() == i){
                flag = 0;
                s.pop();
                continue;
            }
            ans = 0;
            break;
        }
    }
    if (ans)
        cout<<"YES"<<endl;
    else    
        cout<<"NO"<<endl;
}
int main(){
    fast;
    bool test_cases = true;
    if (test_cases){
        int t;
        cin>>t;
        lf(i, 0, t)
            solve();
    }
    else    
        solve();
    return 0;
}