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.
/*
*   Flip_the_Bits.cpp
*   This Code Was Made By : AbDoO_
*   ---------------------------------
*   Created : 2021-04-03 18:11:24.063445
*   Time Taken: 41m
*   --------------------------------
*   CodeForce : codeforces.com/contest/1504/problem/B
*/

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

void io_system(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt" ,"r",stdin);
        freopen("output.txt","w",stdout);
    #endif
}
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
#define fr(a,b) for(int i = a; i < b; i++)
#define rep(i,a,b) for(int i = a; i < b; i++)
#define PAUSE system("PAUSE");
#define ll long long
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).end(),(x).begin()
#define sz(x) (int((x).size()))
#define F first
#define S second
#define read(x) int x;scanf("%d",&x);
#define pf printf
#define print printf
#define L <<'\n'
// ---------------------------------------------------------------------------------
// ---------------------------------------------------------------------------------

string solution(){
    read(n);
    string a,b;
    cin>>a>>b;

    int ons[n];

    ons[0] = a[0] - '0';

    fr(1,n){
        ons[i] = ons[i-1] + (a[i]-'0');
    } 
    
    int flips =0;
    for( int i =n-1 ; i+1 !=0 ; i--){
        int ones,zeros;
        ones = ons[i];
        zeros = i+1 - ons[i];
        if ( (abs((flips%2)*1 - (a[i]-'0') )) != b[i]-'0' ){
            if(ones != zeros ){
                //pf("%d|%d  - %d",ones,zeros,flips);
                return "NO";
            }
            else{
                flips++;
            }
        }
        //rep(ii,0,n)
            //pf("%d ",(abs((flips%2)*1 - (a[ii]-'0'))));
        //cout L;

    }
    return "YES";

}

int main(){
    io_system();
    int n_testcase = 1;
    cin >> n_testcase;  
    while(n_testcase--){
        cout<< solution() L;
    }
}


// ICPC Helper | By: Abdoo
/*        if(abs((flips%2)*1){
            ones = n - ons[i];
            zeros =  ons[i];
        }
        else{
            ones = ons[i];
            zeros = n - ons[i];
        }
*/