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;
int CalcSumOfDigit(int n);
string upper(string str);
string lower(string str);
const int INF = 1e9;
int main() {
  string a, b, c;
  if ((a[int((a).size())] == b[0]) && (b[int((b).size())] == c[0]))
    cout << "YES" << endl;
  else {
    cout << "NO" << endl;
  }
  return 0;
}
int CalcSumOfDigit(int n) {
  int s = 0;
  while (n) {
    s += n % 10;
    n = n / 10;
  }
  return s;
}
string upper(string str) {
  for (auto itr = str.begin(); itr != str.end(); itr++) {
    if (97 <= *itr && *itr <= 122) {
      *itr = *itr - 32;
    }
  }
  return str;
}
string lower(string str) {
  for (auto itr = str.begin(); itr != str.end(); itr++) {
    if (65 <= *itr && *itr <= 90) {
      *itr = *itr + 32;
    }
  }
  return str;
}
