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;
string divide[4] = {"dream", "dreamer", "erase", "eraser"};
int main() {
  int N, C, K;
  cin >> N;
  vector<int> T(N);
  for (int i = 0; i < N; i++) {
    cin >> T.at(i);
  }
  long sum = 0;
  long cnt1 = 0;
  for (int i = 0; i < N; i++) {
    sum += T.at(i);
    if (i % 2 == 0) {
      if (sum <= 0) {
        cnt1 += -sum + 1;
        sum = 1;
      }
    } else {
      if (sum >= 0) {
        cnt1 += sum + 1;
        sum = -1;
      }
    }
  }
  long cnt2 = 0;
  sum = 0;
  for (int i = 0; i < N; i++) {
    sum += T.at(i);
    if (i % 2 == 1) {
      if (sum <= 0) {
        cnt2 += -sum + 1;
        sum = 1;
      }
    } else {
      if (sum >= 0) {
        cnt2 += sum + 1;
        sum = -1;
      }
    }
  }
  cout << min(cnt1, cnt2) << endl;
  return 0;
}
