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 n, W;
long long w[100], v[100];
long long dp[101][950001];
int main() {
  cin >> n >> W;
  for (int i = 0; i < n; i++) {
    cin >> w[i] >> v[i];
  }
  long long wsum = 0;
  for (int i = 0; i < n; i++) wsum += w[i];
  if (wsum <= W) {
    long long vsum = 0;
    for (int i = 0; i < n; i++) vsum += v[i];
    cout << vsum << endl;
  } else if (w[0] > W) {
    cout << 0 << endl;
  } else {
    if (w[0] > 30) {
      long long basis = w[0];
      long long plus = 401;
      long long c = min(101LL, W / basis);
      W = W - basis * c + plus * c;
      for (int i = 0; i < n; i++) w[i] = w[i] - basis + plus;
    }
    for (int i = 0; i < n + 1; i++)
      for (int j = 0; j < W + 1; j++) dp[i][j] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < W; j++) {
        if (j - w[i] + 1 >= 0) {
          dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i][j - w[i] + 1] + v[i]);
        } else {
          dp[i + 1][j + 1] = dp[i][j + 1];
        }
      }
    }
    cout << dp[n][W] << endl;
  }
}
