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;
const int maxn = 100;
long long N, W;
long long w[maxn + 1];
long long v[maxn + 1];
long long cnt[maxn + 1][500] = {0};
long long dp[maxn + 1][500] = {0};
long long ans = 0;
long long b = 0;
int main() {
  cin >> N >> W;
  for (int i = 0; i < N; i++) cin >> w[i + 1] >> v[i + 1];
  b = w[1] - 1;
  for (int i = 1; i <= N; i++) w[i] -= b;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 4 * N; j++) {
      if (w[i] <= j && j <= W - b * (cnt[i - 1][j - w[i]] + 1)) {
        if (dp[i][j] < dp[i - 1][j - w[i]] + v[i]) {
          dp[i][j] = dp[i - 1][j - w[i]] + v[i];
          cnt[i][j] = cnt[i - 1][j - w[i]] + 1;
        }
      }
      if (dp[i][j] < dp[i - 1][j]) {
        dp[i][j] = dp[i - 1][j];
        cnt[i][j] = cnt[i - 1][j];
      }
    }
  }
  for (int i = 0; i <= 4 * N; i++) {
    if (i <= W - b * cnt[N][i]) ans = max(ans, dp[N][i]);
  }
  cout << ans << endl;
  return 0;
}
