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>
using namespace std;

#define ll          long long
#define pb          push_back
#define pii         pair<int,int>
#define pll         pair<long,long>
#define vi          vector<int>
#define vll         vector<ll>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define all(a)      (a).begin(),(a).end()
#define x           first
#define y           second
#define sz(x)       (int)x.size()
#define endl        '\n'
#define mod        1000000007
#define rep(i,a,b)  for(int i=a;i<b;i++)
#define rem(i,a,b)  for(int i=a;i>b;i--)
#define mp(a,b)     make_pair(a,b)
#define vvi(a,b,name)    vector<vector<int>> name(a,vector<int>(b,-1))


void solve(){
    string s1,s2,s3;cin>>s1>>s2>>s3;
    if(s1[s1.length()-1]==s2[0] && s2[s2.length()-1]==s3[0])cout<<"YES";
    else cout<<"NO";

}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
