Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

rng gorilla apple

Sample Output 1

YES

They form a word chain.

Sample Input 2

yakiniku unagi sushi

Sample Output 2

NO

A and B form a word chain, but B and C do not.

Sample Input 3

a a a

Sample Output 3

YES

Sample Input 4

aaaaaaaaab aaaaaaaaaa aaaaaaaaab

Sample Output 4

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <iterator>  // std::back_inserter()
#include <algorithm>  // std::copy()
#include <functional>  // std::greater<T>()
#include <utility>  // std::swap()
#include <numeric>  // accumulate(ALL(vec), 0)  0 は初期値
#include <cmath>
#include <climits>  // INT_MIN
#include <cctype>  // std::isdigit()
using namespace std;

#define ALL(obj) (obj).begin(), (obj).end()
#define REP(i, n) for (int i=0; i < (int)(n); i++)  // 0 ~ n-1
#define REPN(i, n) for (int i=1; i <= (int)(n); i++)  // 1 ~ n
#define DESCEND greater<int>()  // sort 降順 大きい順
#define UNIQUE(vec) (vec).erase(unique(ALL((vec))), (vec).end());  // 必ずsortを行なってから
#define MIN(vec) min_element(ALL((vec)))  // イテレータのため、値を取り出すときは * を先頭につける
#define MAX(vec) max_element(ALL((vec)))
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define SUM(vec) accumulate(ALL((vec)), 0)  // 0 は初期値
#define COPY(vec1, vec2) copy(ALL(vec1), back_inserter(vec2))  // vec1をvec2にコピーする vec2は空にしておく必要あり
typedef long long ll;

const int MOD = 1000000007;  // 1 000 000 007


int main() {
    cin.tie();
    ios::sync_with_stdio(false);

    // input
    string A, B, C;
    cin >> A >> B >> C;
    
    // calculation
    if (A.back() == B.front() && B.back() == C.front()) {
        cout << "YES" << "\n";
    } else{
        cout << "NO" << "\n";
    }
    
    return 0;
}