You are given three strings A, B and C. Check whether they form a word chain.

More formally, determine whether both of the following are true:

* The last character in A and the initial character in B are the same.
* The last character in B and the initial character in C are the same.



If both are true, print `YES`. Otherwise, print `NO`.

Constraints

* A, B and C are all composed of lowercase English letters (`a` - `z`).
* 1 ≤ |A|, |B|, |C| ≤ 10, where |A|, |B| and |C| are the lengths of A, B and C, respectively.

Input

Input is given from Standard Input in the following format:


A B C


Output

Print `YES` or `NO`.

Examples

Input

rng gorilla apple


Output

YES


Input

yakiniku unagi sushi


Output

NO


Input

a a a


Output

YES


Input

aaaaaaaaab aaaaaaaaaa aaaaaaaaab


Output

NO
#include <bits/stdc++.h>
#include <stdio.h>
#include <numeric>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define repone(i,n) for(int i=1;i<(int)(n);++i)
#define repo(i,o,n) for(int i=o;i<(int)(n);++i)
#define Rep(j,n) for(int j=0;j<(int)(n);++j)
#define Repo(j,o,n) for(int j=o;j<(int)(n);++j)
#define Repone(j,n) for(int j=1;j<(int)(n);++j)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define int long long
#define pb(a) push_back(a)
#define eraser(v,i) (v).erase(v.begin()+(i))
#define pbpair(a,b) push_back(make_pair(a,b))
#define MOD 1000000007
#define INF 9223372036854775807
#define pairint pair<int,int>
int num[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
bool uru(int checkyear){if(checkyear%4==0){if(checkyear%100!=0||checkyear%400==0)return true;}return false;}
int gcd( int x, int y ){return y ? gcd(y, x%y) : x;}
signed main() {
    int x=0,y=0,z=0,w=0,n=0,m=0;
    string s,d,f;
cin>>s>>d>>f;
if(s[s.size()-1]==d[0]&&d[size()-1]==f[0])cout << "YES";
    else cout<< "NO";
}
