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 main() {
  uint32_t N, W;
  cin >> N >> W;
  vector<uint32_t> v_w(N, 0), v_v(N, 0);
  for (uint32_t n = 0; n < N; ++n) cin >> v_w[n] >> v_v[n];
  uint32_t dp[N + 1][W + 1];
  for (uint32_t n = 0; n < N + 1; ++n) dp[n][0] = 0;
  for (uint32_t w = 0; w < W + 1; ++w) dp[0][w] = 0;
  for (uint32_t n = 1; n <= N; ++n) {
    for (uint32_t w = 1; w <= W; ++w) {
      if (w >= v_w[n - 1])
        dp[n][w] = max(dp[n - 1][w - v_w[n - 1]] + v_v[n - 1], dp[n - 1][w]);
      else
        dp[n][w] = dp[n - 1][w];
    }
  }
  cout << dp[N][W] << endl;
  return 0;
}
