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.
/**
 *   author:  Niloy_Das_19
 *   created: 2021-04-03 21:52:31 
**/

#include <bits/stdc++.h>
using namespace std;

#define fo(n) for (int i = 0; i < n; i++)
#define F first
#define S second
#define ll long long
#define pb push_back
#define mp make_pair
#define need                     \
    ios::sync_with_stdio(false); \
    cin.tie(NULL);               \
    cout.tie(NULL);
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

int main()
{
    // need
    int t = 1;
    cin >> t;
    getchar();
    while (t--)
    {
        string n, a, b;
        cin >> n >> a >> b;
        if (a == b)
        {
            cout << "YES\n";
            continue;
        }
        int one = 0, zero = 0, one1 = 0, zero1 = 0, i = a.size()-1, c = 0;
        while (i >= 0)
        {
            if (a[i] == b[i])
            {
                i--;
                continue;
            }
            else
            {
                for (int j = i; j >= 0; j--)
                {
                    if (a[j] != b[j])
                    {
                        if (one1 == zero1)
                        {
                            one1 = 0;
                            zero1 = 0;
                            if (a[j] == '1')
                                one++;
                            else
                                zero++;
                        }
                        else
                        {
                            c = 1;
                            i = -1;
                            break;
                        }
                    }
                    else
                    {
                        if (one == zero)
                        {
                            one = 0;
                            zero = 0;
                            if (a[j] == '1')
                                one1++;
                            else
                                zero1++;
                        }
                        else
                        {
                            c = 1;
                            i = -1;
                            break;
                        }
                    }
                }
            }
            i = -1;
        }
        if (c||one!=zero||one1!=zero1)
            cout << "NO\n";
        else
            cout << "YES\n";
    }
    return 0;
}