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.
// #pragma GCC optimize(3, "Ofast", "inline")

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<double, double> Pd;
typedef pair<ll, ll> pll;

template <class T>
inline void read(T &x)
{
    x = 0;
    int sign = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            sign = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    x = x * sign;
}

template <class T>
inline void wirte(T x)
{
    if (x < 0)
    {
        x = -x;
        putchar('-');
    }
    if (x >= 10)
        wirte(x / 10);
    putchar(x % 10 + '0');
}

const int INF = 0x3f3f3f3f;
const ll LINF = 2e18;
const int maxn = 3e5+5;
const int maxm = 1e4+5;
const ll mod = 1e9+7;
const double PI = acos(-1);
const double eps = 1e-7;

int T,n;
char a[maxn],b[maxn];
int suma[maxn];

int main()
{
#ifndef ONLINE_JUDGE
    freopen(".in", "r", stdin);
    // freopen(".out","w",stdout);
#endif
    read(T);
    while (T--)
    {
        read(n);
        scanf("%s%s",a+1,b+1);
        for(int i=1;i<=n;i++){
            suma[i] = suma[i-1] + ((a[i]=='1') ? 1 : -1);
        }
        bool flag=true;
        int cnt=0;
        for(int i=n;i;i--){
            int na = (a[i] - '0' + cnt)%2 , nb = b[i]-'0';
            if(na==nb) continue;
            if(suma[i]){
                flag=false;
                break;
            }
            cnt++;
        }
        if(flag) puts("YES");
        else puts("NO");
    }
    
    
    return 0;
}