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>
typedef long long ll;
typedef long double ld;
 
#define f(i,a,b) for(int i = a;i<b;i++)
#define rep(i,a,b) for(int i = a;i<=b;i++)
#define fd(i,a,b) for(int i = a;i>b;i--)
#define repd(i,a,b) for(int i = a;i>=b;i--)
 
#define pii pair<int,int>
#define pll pair<ll,ll>
 
#define fi first
#define se second
#define pb push_back
#define sz(x) (int) (x).size()
#define all(x) x.begin(),x.end()
#define endl '\n'
using namespace std;
const int mod = 1000000007;
const int maxn = 2e5;

int main(){
    int t;cin>>t;
    while(t--){
        int n;cin>>n;
        string a,b;
        cin>>a>>b;
        int one = 0,zero = 0, inv = 0;
        f(i,0,n){
            if(a[i] == '0') ++zero;
            else ++one;
        }
        bool ans = true;
        repd(i,n-1,0){
            char curr = (inv&1)?!(a[i]-'0') + '0':a[i];
            if(curr != b[i]){
                if(i%2 == 0){
                    ans = false;
                    break;
                }
                else{
                    if(one != zero){
                        ans = false;
                        break;
                    }
                    ++inv;
                    swap(zero,one);
                    if(curr == '1') --zero;
                    else --one;
                }
            }
            else{
                if(curr == '1') --one;
                else --zero;
            }
        }
        if(ans) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;

    }

}