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, w[111], v[111], dp[111111], ans;
vector<pair<int, int> > vec;
int main() {
  cin >> N >> W;
  for (int i = 1; i <= N; ++i) scanf("%d%d", &w[i], &v[i]);
  if (w[1] > 400) {
    for (int i = 1; i <= N; ++i) vec.push_back(make_pair(v[i], -w[i]));
    sort(vec.begin(), vec.end());
    for (int i = N - 1; i >= 0; --i)
      if (W >= (-vec[i].second)) ans += vec[i].first, W += vec[i].second;
    cout << ans << endl;
    return 0;
  }
  W = min(W, 43333);
  for (int i = 1; i <= N; ++i)
    for (int j = W; j >= w[i]; --j) dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
  for (int i = 1; i <= W; ++i) ans = max(ans, dp[i]);
  cout << ans << endl;
  return 0;
}
