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;
template <typename T>
long long sz(vector<T> &vec) {
  return (long long)vec.size();
}
long long sz(string &s) { return (long long)s.size(); }
long long gcd(long long a, long long b) { return ((!b) ? a : gcd(b, a % b)); }
long long dp[101][101][301] = {};
int main() {
  long long n, W;
  cin >> n >> W;
  long long w[n], v[n];
  long long wmemo = 0;
  for (long long i = 0; i < n; i++) {
    cin >> w[i] >> v[i];
    if (i == 0) wmemo = w[i];
    w[i] -= wmemo;
  }
  for (long long i = 1; i < n + 1; i++) {
    for (long long j = 1; j < n + 1; j++) {
      for (long long k = 0; k < 301; k++) {
        dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k]);
        if (k - w[i - 1] >= 0)
          dp[i][j][k] =
              max(dp[i][j][k], dp[i - 1][j - 1][k - w[i - 1]] + v[i - 1]);
      }
    }
  }
  long long ans = 0;
  for (long long j = 1; j < n + 1; j++) {
    for (long long k = 0; k < 301; k++) {
      if (j * wmemo + k <= W) ans = max(ans, dp[n][j][k]);
    }
  }
  cout << ans << endl;
  return 0;
}
