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.
#ifdef IVELOID
 #define debug(x) cerr<<#x<<":"<<x<<endl
#else
 #define debug(x)
#endif
#include<bits/stdc++.h>
using namespace std;
#define int long long
using ll=long long;
using ld=long double;
const int mod=998244353;
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=(a);i<=b;++i)
#define inc(i,a,b) for(int i=(a);i<(int)b;++i)
using pii=pair<int,int>;
#define fi first
#define se second
using vi=vector<int>;
using vvi=vector<vector<int>>;
using vpii=vector<pii>;
template<class T>T cmax(T &a,T b){return a=max(a,b);}
template<class T>T cmin(T &a,T b){return a=min(a,b);}
ll acc(const vector<int> &a){return accumulate(all(a),0LL);}
double acc(const vector<double> &a){return accumulate(all(a),0.0);}
ll qpow(ll a,ll x,ll p=mod){ll res=1%p;a=a%p;for(;x;x>>=1,a=a*a%p) if(x&1) res=(res*a)%p;return res;}

const int inf=1e9;
// head

void solve(){
  int n;
  string s,t;
  cin>>n;
  cin>>s>>t;
  vi pre(n);
  int c0=0,c1=0;
  for(int i=0;i<n;++i){
    c0+=(s[i]=='0');
    c1+=(s[i]=='1');
    pre[i]=(c0==c1);
  }
  int cur=0;
  vi a(n),b(n);
  inc(i,0,n) a[i]=s[i]-'a';
  inc(i,0,n) b[i]=t[i]-'a';
  int ok=1;
  for(int i=n-1;i>=0;--i){
    if((a[i]+b[i]+cur)%2==0) continue;
    else{
      if(pre[i]==true){
        cur^=1;
      }
      else{
        ok=0;
      }
    }
  }
  cout<<(ok?"YES\n":"NO\n");
}

signed main() {
  ios::sync_with_stdio(false),cin.tie(0);
  int t;
  cin>>t;
  while(t--) solve();
  return 0;
}
