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;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
const int INF = 1e18 + 9;
constexpr long double pi = 3.141592653589793238462643383279;
int fact(int i) {
  if (i == 0) return 1;
  return (fact(i - 1)) * i;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int keta(int n) {
  if (n == 0) return 1;
  int count = 0;
  while (n != 0) {
    n /= 10;
    count++;
  }
  return count;
}
int ketasum(int n) {
  int sum = 0;
  while (n != 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
string a, b, c;
int main(void) {
  cin >> a >> b >> c;
  (a[a.size() - 1] == b[0] && b[b.size() - 1] == c[0] ? cout << "NO" << endl
                                                      : cout << "YES" << endl);
  return 0;
}
