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<iostream>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<iomanip>
#include<stack>
#include<cmath>

 
using namespace std;
 
 
#ifndef POJ
    #define Cpp11
#endif
 
#ifdef Cpp11
    #define _for(i,a,b) for(auto i = (a);i<(b);++i)
    #define _rep(i,a,b) for(auto i = (a);i<=(b);++i)
    #define _dep(i,a,b) for(auto i = (a);i>=(b);--i)
    #define mt(a,b,c) make_tuple(a,b,c) 
    #include<ctime>
    #include<random>
    #include<chrono>
    #include<limits>
    #include<complex>
    #include<unordered_set>
    #include<unordered_map>
    #include<cassert>
    #include<functional>
    #include<bitset>
    #define IO ios::sync_with_stdio(false);cout.tie(nullptr);cin.tie(nullptr);
#endif
 
#ifndef Cpp11
    #define _for(i,a,b) for(int i = (a);i<(b);++i)
    #define _rep(i,a,b) for(int i = (a);i<=(b);++i)
    #define _dep(i,a,b) for(int i = (a);i>=(b);--i)
    #define IO ios::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);
#endif
 
 
using namespace std;


 
#define fr first
#define sc second
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define sci(x) scanf("%d",&(x))
#define scl(x) scanf("%lld",&(x))
#define debugline fprintf(stderr,"-----------------------\n");
#define szint(x) ((int)(x).size())


 
#ifdef I128IO
typedef __int128 SPLL;
char buff[200];
template<typename T>
void read(T& o)
{
    o = 0;
    scanf("%s",buff);
    //printf("%s\n",buff);
    T f = 1;
    int i = 0;
 
    if(buff[0]=='-')i++,f = -1;
    for(;buff[i];++i)
    {
        o*=10,o+=buff[i]-'0';
    }
    o*=f;
}
 
template<typename T>
void write(T o)
{
    int len = 0;
    buff[0] = '0';
    if(o==0)
    {
        putchar('0');
        return;
    }
    int f = 1;
    if(o<0) f = -1;
    o*=f;
    for(;o;o/=10) buff[len++] = o%10+'0';
    if(f==-1) buff[len++] = '-';
    reverse(buff,buff+len);
    buff[len++] ='\0';
    printf("%s",buff);
}
#endif
 
#ifdef FIO
char buff[100];
template<typename T>
void read(T& o)
{
    o = 0;
    scanf("%s",buff);
    //printf("%s\n",buff);
    T f = 1;
    int i = 0;
    if(buff[0]=='-')i++,f = -1;
    for(;buff[i];++i)
    {
        o*=10,o+=buff[i]-'0';
    }
    o*=f;
}
template<typename T>
void write(T o)
{
    int len = 0;
    buff[0] = '0';
    if(o==0)
    {
        putchar('0');
        return;
    }
    int f = 1;
    if(o<0) f = -1;
    o*=f;
    for(;o;o/=10) buff[len++] = o%10+'0';
    if(f==-1) buff[len++] = '-';
    reverse(buff,buff+len);
    buff[len++] ='\0';
    printf("%s",buff);
}
#endif

#define OIreader


#ifdef OIreader
inline int read(){
    char ch; bool sign=0; int res=0;
    while(!isdigit(ch=getchar()))if(ch=='-')sign=1;
    while(isdigit(ch))res=res*10+(ch^48),ch=getchar();
    if(sign)return -res; else return res;
}
#endif

#define _Random_

#ifdef _Random_
    mt19937 __MT(chrono::system_clock::now().time_since_epoch().count());
#endif
 
#ifndef BigNumber
    typedef unsigned long long uLL;
    typedef long long LL;
    typedef long long ll;
    typedef pair<LL,LL> pLL;
    typedef vector<LL> vL;
    typedef vector<vL> vLL;
#endif

typedef vector<int> vi;
typedef vector<vi > vii;
typedef pair<int,int> pii;


const int maxn = 300000+10;
const int apsz = 26;
const int mod = 1000000000+7;

char s[maxn];
int a[maxn],b[maxn],c[maxn],n;



inline int solve()
{
    set<int> pos;
    pos.insert(-1);
    int now = 0;
    for(int i = 0;i<n;++i)
    {
        c[i] = a[i]^b[i];
        if(a[i]==0)now--;
        else now++;
        if(now==0)
        {
            pos.insert(i);
            //cerr<<i<<endl;
        }
    }
    int i;
    for(i = 0;i<n&&c[i]==0;)++i;
    for(;i<n;)
    {
        int j = i+1;
        while(j<n&&c[j]==1)++j;
        if(!pos.count(j-1)||!pos.count(i-1))return 0;
        i = j;
        while(i<n&&c[i]==0)++i;
    }
    return 1;
}


int main(void)
{
    int T;
    sci(T);
    for(;T--;)
    {
        sci(n);
        scanf("%s",s);
        _for(i,0,n) a[i] = s[i]-'0';
        scanf("%s",s);
        _for(i,0,n) b[i] = s[i]-'0';
        printf("%s\n",solve()?"YES":"NO");
    }
    return 0;
}