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; //int:2*10**9
typedef long double ld;
typedef pair<ll,ll> P;
#define REP(i,n) for(ll i = 0; i<(ll)(n); i++)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define FORD(i,a,b) for(ll i=(a);i>=(b);i--)
#define pb push_back
#define MOD 1000000007 //998244353
#define PI 3.141592653
#define INF 100000000000000 //14
//cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

void solve() {
    ll n; cin >> n;
    string s, t; cin >> s >> t;
    ll ocnt = 0; ll zcnt = 0;
    vector<ll> lines = {0}; ll last = 0;
    REP(i,n) {
        if (s[i]=='1') ocnt++;
        else zcnt++;
        if (ocnt==zcnt) {
            bool jud = true;
            ll checker = 0;
            FOR(j,last,i) {
                if (j==last) {
                    if (s[last]==t[last]) checker=0;
                    else checker=1;
                }
                else {
                    if (checker && s[j]==t[j]) jud = false;
                    else if (checker==0 && s[j]!=t[j]) jud = false;
                }
            }
            if (jud==false) {
                cout << "NO" << endl;
                return;
            }
            last = i+1;
        }
    }
    FOR(i,last,n-1) {
        if (s[i]!=t[i]) {
            cout << "NO" << endl;
            return;
        }
    }
    cout << "YES" << endl;
    return;
}

int main(){
    ll t; cin >> t;
    REP(i,t) solve();
    return 0;
}