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 dp[105][105][500];
int main() {
  ios::sync_with_stdio(false);
  int N, W;
  int w[105];
  int v[105];
  cin >> N >> W;
  for (int i = 1; i <= N; i++) cin >> w[i] >> v[i];
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= i; j++) {
      for (int k = 0; k <= 300; k++) {
        dp[i][j][k] = dp[i - 1][j][k];
        if (k - (w[i] - w[1]) >= 0) {
          dp[i][j][k] =
              max(dp[i - 1][j][k], dp[i - 1][j - 1][k - (w[i] - w[1])] + v[i]);
        }
      }
    }
  }
  int ans = -1;
  for (int i = 1; i <= N; i++)
    if (W - (long long)i * w[1] >= 0) ans = max(ans, dp[N][i][W - i * w[1]]);
  cout << ans << endl;
  return 0;
}
