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.
#define watch5(a,b,c,d,e) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d << " | "<<#e<<": "<<e<<endl;
#define watch4(a,b,c,d) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
#define watch3(a,b,c) cerr <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<endl;
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define sharingan cerr << "\n-------------------------------------------\n\n";
#define watch2(a,b) cerr <<#a<<": "<<a<<" | "<<#b<<": "<<b<< endl;
#define RUN ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define watch(a) cerr <<#a<<": "<<a<<endl;
#define __gcd(a, b) __algo_gcd(a, b)
#define PI 3.14159265358979323846
#define mod 1000000007
#include <algorithm>
#include <bitset>
#include <deque>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <new>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <valarray>
#include <vector>
#include <array>
#include <random>
#include<cstring>
#include <unordered_map>
#include <unordered_set>
#define int long long
using namespace std;

bool pal( string s ){
    string z = "";
    z += s;
    reverse(s.begin(), s.end());
    return (z == s);
}

int32_t main(){
    int t;
    cin >> t;
    while( t-- ){
        int n;
        cin >> n;
        string a, b;
        cin >> a >> b;
        int same = 0, dif = 0, pre = 1;
        int o = 0, z = 0;
        int bo = 0, bz = 0;
        for( int i = 0; i < n; i++ ){
            o += (a[i] == '1');
            z += (a[i] == '0');
            if( a[i] == b[i] ){
                same = 1;
            } 
            else{
                dif = 1;
            }
            if( o == z ){
                if( same == 1 && dif == 1 ){
                    pre = 0;
                    break;
                }
                same = 0;
                dif = 0;
            }
        }
        if( pre == 1 && dif == 0 ){
            cout << "YES\n";
        }
        else{
            cout << "NO\n";
        }
    }
}