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>

#define x first
#define y second
#define pb push_back
#define pii pair<int, int>
#define pll pair<ll, ll>
#define ll long long
#define mp make_pair
#define ld long double

using namespace std;

const double eps = 1e-9, pi = 3.1415;
const ll INF = 1e18, LINF = 1e9;
const int maxn = 1e6, mod = 1e9 + 7, block = 350;

void file_in_out(){
   // freopen("angle1.in","r",stdin);
   // freopen("angle1.out","w",stdout);
}

int t, n;
string a, b;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--){
        cin >> n;
        cin >> a >> b;
        int k = 0, k2 = 0;
        for (auto c : a){
            k += (c == '1');
            k2 += (c == '0');
        }
        int cur = 0;
        bool ok = 0;
        for (int i = n - 1; i >= 0 && !ok; i--){
            if (cur == 0){
                if (a[i] != b[i]){
                    if (k != k2)
                        ok = 1;
                    else{
                        cur ^= 1;
                        if (a[i] == '0')
                            k--;
                        else
                            k2--;
                    }
                }
                else{
                    k -= (a[i] == '1');
                    k2 -= (a[i] == '0');
                }
            }
            else{
                if (a[i] == b[i]){
                    if (k != k2)
                        ok = 1;
                    else {
                        cur ^= 1;
                        if (a[i] == '0')
                            k2--;
                        else
                            k--;
                    }
                }
                else{
                    k -= (a[i] == '0');
                    k2 -= (a[i] == '1');
                }
            }
         //   cout << i << " " << ok << '\n';
        }
        if (ok){
            cout << "NO\n";
        }
        else{
            cout << "YES\n";
        }
    }
}
/*
*/
