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.
// stryker27

#include<bits/stdc++.h>
using namespace std;

#define  F           first
#define  S           second
#define  mp          make_pair
#define  pb          push_back
#define  int         long long int
#define  vi          vector<int>
#define  vpi         vector<pair<int,int>>

#define  For(b)     for(int i=0;i<b;i++)
#define  f(i,a,b) for(int i=a;i<b;i++)

#define  nl          "\n"
#define  gr         greater<int>()
#define  all(x)     x.begin(),x.end()
#define  flash      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
const int mod  =    1000000007;

void print()  {    cout << endl;  }
void printarray()  {    return;  }
void scan() { return; }

template<typename T , typename... Args>  void print(T a , Args... arg)    {    cout << a << " " ;    print(arg...);    }

template<typename T , typename... Args> void scan(T &a , Args&... arg) {   cin >> a;   scan(arg...);   }

template<typename T, typename... Args> void scan(vector<T>&v , Args&... arg) {  for (auto &i : v)  cin >> i;  scan(arg...);  }

template<typename T, typename... Args> void printarray(vector<T>&v, Args&... arg) {   for (auto i : v)     cout << i << " ";   cout << endl;    printarray(arg...);    }

const int mxN = 1e6 + 5;
vi prime(mxN);

void sieve()
{
    for (int i = 0; i < mxN; i++)
    {
        prime[i] = true;
    }
    prime[0] = false;
    prime[1] = false;

    for (int p = 2; p * p < mxN; p++)
    {
        if (prime[p])
        {
            for (int i = p * p; i < mxN; i += p)
            {
                prime[i] = false;
            }
        }
    }
}


bool palindrome(const string &s) {
    int n = s.length();
    for (int i = 0; i < n; i++) {
        if (s[i] != s[n - i - 1]) return false;
    }
    return true;
}

int isprime(int x) {

    for (int i = 2; i <= x / 2; i++)
    {
        if (x % i == 0) {

            return 0;
        }

    }
    return 1;

}

int power(int x, unsigned int y)
{
    if ( y == 0)
        return 1;
    if (y % 2 == 0)
        return power(x, y / 2) * power(x, y / 2);
    return x * power(x, y / 2) * power(x, y / 2);
}

int gcd(int a , int b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int lcm (int a , int b) {
    return a * b / gcd(a, b);
}

// ================= Answer =================

void stryker27()
{

    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    int j = 0, one = 0, zero = 0;
    vector<pair<int, int>>x;
    for (int i = 0; i < n; i++)
    {
        if (a[i] == '1') one++;
        else zero++;
        if (one == zero) {
            x.push_back({j, i});
            j = i + 1;
        }
    }
    for (auto i : x)
    {
        int start = i.first;
        int end = i.second;
        if (a[start] == b[start]) continue;
        else {
            for (int j = start; j <= end; j++) {
                if (a[j] == '0') a[j] = '1';
                else a[j] = '0';
            }
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (a[i] != b[i]) {
            cout << "NO" << "\n";
            return;
        }
    }
    cout << "YES" << "\n";
}


// ============== X Answer X ================

int32_t main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    flash;
    //sieve();
    int t;
    cin >> t;
    while (t--)
        stryker27();

    return 0;
}


