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;
#define fastio ios_base::sync_with_stdio(false); cin.tie(0) ; cout.tie(0);
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;i<=n;i++)
#define ll long long
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define mii map<ll,ll>
#define pii pair<ll,ll>
#define vi vector<ll>
#define endl "\n"
#define deb1(x)                cout<<#x<<": "<<x<<endl
#define deb2(x, y)             cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define deb3(x, y, z)          cout<<#x<<":"<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace1(x)                cerr<<#x<<": "<<x<<endl
#define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<" : "<<f<<endl
const ll mod = 1e9+7;
#define popcount(x) __builtin_popcount(x)
#define all(x) x.begin(),x.end()
#define mem(a,v) memset(a,v,sizeof(a))
#define Max(x,y,z) max(x,max(y,z))
#define Min(x,y,z) min(x,min(y,z))

ll dx[]={-1,0,1,0};
ll dy[]={0,-1,0,1};

ll msum(ll a,ll b,ll m) { return (a%m+b%m)%m; }
ll msub(ll a,ll b,ll m) { return (a%m-b%m)%m; }
ll mpro(ll a,ll b,ll m) { return ((a%m)*(b%m))%m; }
ll m_m(ll a,ll b,ll m);
ll fxp(ll a,ll b,ll m);
void swap(ll &a,ll &b){ ll t=a; a=b; b=t;}

void solve()
{
    ll n,i;
    cin>>n;
    string s1,s2;
    cin>>s1>>s2;
    ll flag=0;
    i=0;
    ll one=0,zero=0;
    while(i<n)
    {
        if(s1[i]!=s2[i])
        {
            ll j=i;
            ll tzero=0,tone=0;
            while(j<n && s1[j]!=s2[j])
            {
                if(s1[j]=='1')
                {
                    one++;
                    tzero++;
                }
                else    
                {
                    zero++;
                    tone++;
                }

                j++;
            }

            if(one==zero && tzero==tone)
            {
                ll k;
                for(k=i;k<j;k++)
                {
                    s1[k]=s2[k];
                }
                i=j;
            }
            else
            {
                // trace4(i,j,one,zero);
                flag=1;
                break;
            }

        }
        else
        {
            if(s1[i]=='1')
                one++;
            else
                zero++;
            i++;
        }
    }

    if(s1!=s2)
    {
        flag=1;
    }

    if(flag)
    {
        cout<<"NO";
    }
    else
        cout<<"YES";
    cout<<endl;
}

int main()
{
    fastio
    int t=1;
    cin>>t;
    while(t--)
    {
        solve();   
    }
 return 0;
}
ll fxp(ll a,ll b,ll m) {
    if(b==0)
        return 1;
    if(b%2==0)
        return fxp(m_m(a,a,m),b/2,m);
    return m_m(fxp(a,b-1,m),a,m);
}
ll m_m(ll a,ll b,ll m) 
{
    ll res=0;
    a=a%m;
    while(b)
    {
        if(b&1)
        {
            res+=a; 
            res=res%m;
        }
        a=(a*2)%m;
        b=b/2;
    }
    return res;
}