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 int long long
#define all(x) x.begin(),x.end()
#define fi(i,a,b) for(int i=a;i<b;i++)
#define fe(a) for(auto e : a)
#define fr(i,n) for(int i=0;i<n;i++)
#define inarr(arr,n) for(int i=0;i<n;i++)cin>>arr[i]
#define invtr(vtr,n) for(int i=0;i<n;i++){ int x;cin>>x;vtr.push_back(x);}
#define vt(a,n) vector<int> a(n)
#define vtrin(a,n) vt(a,n);inarr(a,n)
#define vtval(a,n,val) vector<int> a(n,val)
#define mod (int)(1e9 + 7)
#define ln "\n"
#define spc " "

using namespace std;

int factorCeil(int a, int b);

int logPower(int x, int y);


string solve()
{
    int n;
    cin>>n;

    string a,b;
    cin>>a>>b;

    vector<bool> changeNeeded(n);

    fi(i,0,n)
    {
        changeNeeded[i] = a[i] != b[i];
    }

    int start = n-1;

    while(start >= 0)
    {
        if(changeNeeded[start])
        {
            break;
        }

        start--;
    }

    if(start < 0)
    {
        return "YES";
    }

    vector<int> zeros(n,0),ones(n,0);

    if(a[0] == '0')
    {
        zeros[0]++;
    }

    else
    {
        ones[0]++;
    }


    fi(i,1,n)
    {
        zeros[i] = zeros[i-1] + (a[i] == '0');
        ones[i] = ones[i-1] + (a[i] == '1');
    }


    for(int i = start; i >= 0; i--)
    {
        if(i == n-1 )
        {
            if(zeros[i] != ones[i])
            {
                return "NO";
            }
        }

        else if(changeNeeded[i] != changeNeeded[i+1])
        {
            if(zeros[i] != ones[i])
            {
                return "NO";
            }
        }
    }

    return "YES";

}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif

    int t;
    cin>>t;

    while(t--)
    {
        cout<<solve()<<ln;
    }

    return 0;
}

int factorCeil(int a, int b)
{
    return (a + b - 1)/b;
}

int logPower(int x, int y)
{
    int res = 1;

    if (x == 0) return 0;

    while (y > 0)
    {
        if (y & 1)
            res = (res*x);

        y = y>>1; // y = y/2
        x = (x*x);
    }
    return res;
}