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;
using ll = long long;
using llu = unsigned long long;
using ld = long double;

#define CEIL(x, y) (ll)ceil((ld)(x) / (ld)(y))

#define dbgPR(ds) cout << ds.first << ' ' << ds.second
#define dbgLST(ds) for(auto it : ds) cout << it << ' '
#define dbgMP(ds) for(auto it : ds) cout << it.first << ' ' << it.second << '\n'
#define print_case cout << "Case " << tc << ": "

const double PI = acos(-1.0), g = 9.80665;
const int unseen = -1;
const ll inf = 1e18 + 505, ninf = - 1e18 - 505;
const int MOD = 1000000007;

void solve(void)
{
    int n;
    string a, b;
    cin >> n >> a >> b;
    vector<vector<int>> freq;
    freq.assign(n, vector<int> (2, 0));

    int cnt0 = 0, cnt1 = 0;
    for(int i = 0; i < a.size(); i++) {
        if(a[i] - '0') cnt1++;
        else cnt0++;
        freq[i][0] = cnt0;
        freq[i][1] = cnt1;
    } 

    bool state = 0;    
    for(int i = a.size() - 1; i >= 0; i--) {
        if(state == 0) {
            if(a[i] != b[i]) {
                if(freq[i][0] != freq[i][1]) {
                    cout << "NO\n";
                    return;
                }
                state = 1;
            }
        }
        else {
            if(a[i] == b[i]) {
                if(freq[i][0] != freq[i][1]) {
                    cout << "NO\n";
                    return;
                }
                state = 0;
            }   
        }
    }

    cout << "YES\n";
}

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(nullptr); 

    int testcase = 1;
    cin >> testcase;
    for(int tc = 1; tc <= testcase; tc++) {
        // print_case;
        solve();
    }  
    return 0;   
}