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 vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using vpii = vector<pii>;
using vs = vector<string>;
using vd = vector<double>;
using ul = unsigned long;
template <class T, class C>
void chmax(T& a, C b) {
  a > b ?: a = b;
}
template <class T, class C>
void chmin(T& a, C b) {
  a < b ?: a = b;
}
const int mod = 1e9 + 7;
struct mint {
  ll x;
  mint(ll x = 0) : x(x % mod) {}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
};
mint c[4005][4005];
void init() {
  c[0][0] = 1;
  for (int i = 0; i <= 4000; i++) {
    for (int j = 0; j <= i; j++) {
      c[i + 1][j] += c[i][j];
      c[i + 1][j + 1] += c[i][j];
    }
  }
}
mint comb(int n, int k) { return c[n][k]; }
bool IsPrime(int num) {
  if (num < 2)
    return false;
  else if (num == 2)
    return true;
  else if (num % 2 == 0)
    return false;
  double sqrtNum = sqrt(num);
  for (int i = 3; i <= sqrtNum; i += 2) {
    if (num % i == 0) {
      return false;
    }
  }
  return true;
}
map<ll, ll> primeFact(ll n) {
  map<ll, ll> res;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      res[i]++;
      n /= i;
    }
  }
  if (n != 1) res[n]++;
  return res;
}
void printv(vi& v) {
  for (int i = (int)(0); i < (int)(v.size()); ++i) cout << v[i] << " ";
  cout << endl;
}
void printv(vll& v) {
  for (int i = (int)(0); i < (int)(v.size()); ++i) cout << v[i] << " ";
  cout << endl;
}
int vtotal(vi& v) {
  int total = 0;
  for (int i = (int)(0); i < (int)(v.size()); ++i) total += v[i];
  return total;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  string a, b, c;
  cin >> a >> b >> c;
  string ans = "No";
  if (a[a.size() - 1] == b[0] && b[b.size() - 1] == c[0]) ans = "Yes";
  cout << ans << endl;
  ;
  return 0;
}
