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
// 失敗するからこそ そこから立ち向かって行く強さがあってそんな強さが本当の強さだと私は思うから
// ぜったいあきらめない
#if defined(DAIJOBU)
#include "/home/ichigo/includes.hpp"
#include "/home/ichigo/debug.hpp"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#include <bits/stdc++.h>
#define deb(x...)
#endif
#define FIO {ios::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);}

using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


int main(){
	FIO
	
	string s1, s2, s3;
	cin >> s1 >> s2 >> s3;
	if(s1[s1.size() - 1] == s2[0] && s2[s2.size() - 1] == s3[0])puts("YES");
	else puts("NO")''
	return 0;
}
