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 <iostream>
#include <vector>
#include <cstdint>
#include <algorithm>

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;
}
    dp[n][0] = 0;

  for (uint32_t w=0; w<W+1; ++w)
    dp[0][w] = 0;
  
  for (uint16_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;
}