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>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef map<int,int> mii;
typedef map<ll,ll> mllll;
typedef map<int, vi> mivi;
typedef map<ll, vll> mllvll;

#define F first
#define S second
#define PB push_back
#define MP make_pair

#define SV(v) sort((v).begin(), (v).end())
#define REP0(a,b) for (int i = (a); i <= (b); ++i)
#define REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPL0(a,b) for (int i = (a); i < (b); ++i)
#define REPL(i,a,b) for (int i = (a); i < (b); ++i)

#define DTI(a) (int)((a)-'0')
#define LATI(a) (int)((a)-'a')
#define UATI(a) (int)((a)-'A')

void read_arri(int &n, vi &arr) {
    int k;
    REPL0(0,n) {
        cin >> k;
        arr.PB(k);
    }
}

void read_arrll(int &n, vll &arr) {
    ll k;
    REPL0(0,n) {
        cin >> k;
        arr.PB(k);
    }
}

//struct P {
//    int x, y;
//    bool operator<(const P &p) {
//        if (x != p.x) return x < p.x;
//        else return y < p.y;
//    }
//};

//bool comp(string a, string b) {
//    if (a.size() != b.size()) return a.size() < b.size();
//    return a < b;
//}
//sort(v.begin(), v.end(), comp);

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

//printf("%.9f\n", x);

// x %= m;
// if (x < 0) x += m;

void solve() {
    int n,i;
    string a,b;
    cin >> n;
    cin >> a;
    cin >> b;
    int cs[n][2];
    memset(cs,0,sizeof(cs));
    (a[0] == '1') ? ++cs[0][1] : ++cs[0][0];
    REPL(i,1,n) {
        (a[i] == '1') ? ++cs[i][1] : ++cs[i][0];
        cs[i][0] += cs[i-1][0];
        cs[i][1] += cs[i-1][1];
    }

    bool isv = true, isf = false;
    for (int j = n-1; j >= 0; --j) {
        if (isf) {
            if (a[j] == b[j]) {
                if (cs[j][0] != cs[j][1]) {
                    isv = false;
                    break;
                }
                isf = false;
            }
        }
        else {
            if (a[j] != b[j]) {
                if (cs[j][0] != cs[j][1]) {
                    isv = false;
                    break;
                }
                isf = true;
            }
        }
    }
    if (isv)  {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int t;
	cin >> t;
	while (t--) {
        solve();
	}

	return 0;
}
