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>
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(NULL)
#ifdef LOCAL
#define READF freopen("in.txt","r",stdin)
#else
#define READF if(0)
#define cerr if(0) cout
#endif
#define fr first
#define sc second
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define chk cerr<<"masuk"<<endl
#define deb(x) cerr << #x << " = " << x <<'\n'
#define arpr(ar_,t_) 	cerr<< #ar_ << ":\n";					\
						for(int i_=0;i_<t_;i_++){				\
							if(i_==t_-1) cerr<<ar_[i_]<<'\n';	\
							else cerr<<ar_[i_]<<" "; 			\
						}
using namespace std;
const ll N = 1e5 + 5;

ll n,m,i,j;
vector<ll> flip;
bool ev;
string a,b;

int main(){
	FASTIO; READF;
	ll t;
	cin>>t;
	while(t--){
		flip.clear();
		cin>>n>>a>>b;
		ev = a[0] == b[0];
		ll ztala = 0, otala = 0, ztalb = 0, otalb = 0;
		i = 0;
		while(i < n){
			ll st = i;
			while((a[st] == b[st]) == (a[i] == b[i]) && i < n){
				a[i] == '0' ? ztala++ : otala++;
				b[i] == '0' ? ztalb++ : otalb++;
				i++;
			}
		//	deb(ztala); deb(otala);deb(ztalb); deb(otalb);
			// flip.pb(i - 1);
			if(otala != otalb || ztala != ztalb) break;
			if(i < n && otala != ztala) break;
			else{
				otala = 0;ztala = 0;
				otalb = 0;ztalb = 0;
			}
		}
		(otala == otalb && ztala == ztalb && otala == ztala) ? cout<<"YES"<<endl : cout<<"NO"<<endl;
		// arpr(flip, flip.size());
	}
}

