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 : Subhrangshu Pandey
* INSTITUTION : HIT, Kolkata
**********************************************/
#include<bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define mod ((long long)(1e9+7))
#define endl "\n"
#define all(x) (x).begin(), (x).end()
//smaller always visited and larger never
#define rep(i, begin, end) for(ll i = begin; i < end; ++i)
#define rev(i, begin, end) for(ll i = begin; i >= end; --i)
#define timetaken cerr<<"Time : "<<1000*(long double)clock()/(long double)CLOCKS_PER_SEC<<"ms\n"
 
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    
	ll t; cin>>t; while(t--){
		ll n;
		cin>>n;
		string s1, s2;
		cin>>s1>>s2;

		bool isposs = true;
		bool allsm = true, allsw = true;
		ll cnt0 = 0, cnt1 = 0;
		rep(i, 0, n){
			if(s1[i] == s2[i])
				allsw = false;
			else allsm = false;
			// if(!(allsm || allsw)){
			// 	isposs = false;
			// 	break;
			// }
			(s1[i] == '1') ? cnt1++ : cnt0++;
			// cout<<i<<" allsw: "<<allsw<<" allsm: "<<allsm<<" cnt0: "<<cnt0<<" cnt1: "<<cnt1<<endl;
			if(cnt1 == cnt0 && (allsw || allsm)){
				allsm = true;
				allsw = true;
				cnt0 = 0, cnt1 = 0;

			// cout<<i<<" in allsw: "<<allsw<<" allsm: "<<allsm<<" cnt0: "<<cnt0<<" cnt1: "<<cnt1<<endl;
			}
			else if(i == n-1 && !allsm){
				isposs = false;
			}
		}
		cout<<(isposs ? "YES\n" : "NO\n");
	}

	timetaken;
	return 0;
}