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
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;
#ifndef FAST_IO
 #define IOS ios_base::sync_with_stdio(false);cin.tie(nullptr);
#endif

#define FILE_IO freopen("input.txt","r",stdin)
#define int long long
#define f first
#define s second
#define endl '\n'
#define all(c) c.begin(),c.end()
#define bp __builtin_popcount
#define debug(x) cerr << #x << " = " << x << endl;

const int MOD = 1e9 + 7;
const int INF = 1e18;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>void print(T t){for(auto &it : t)cout<<it<<" ";}
inline int powerexp(int a, int b, int m = MOD){ int res = 1;while(b){if(b&1)res = res * a % m;a = a * a % m,b >>= 1;}return res % m;}
inline int modinv(int a){ return powerexp(a,MOD-2,MOD);}
inline int lcm(int a, int b){ int g = __gcd(a,b);return a/g*b;}
inline int add(int x, int y){ x += y; if (x >= MOD) x -=  MOD; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % MOD;}
inline int sub(int x, int y){ x -= y; if (x<0) x += MOD; return x ;}

//-------------------< start here >-------------------------// 

void print_vector(vector<int> v){
  for(int i = 0; i < (int)v.size(); i++){
	  if (i == (int)v.size() - 1) cout << v[i];
	  else cout << v[i] << " ";  
	}
  cout << endl;	
}

//char grid[100][100];
void solve(){
   string a, b, c;
   cin >> a >> b >> c;
   bool ok =  (a.end() == b.begin() && b.end() == c.begin());
   if(ok) cout << "YES" << endl;
   else cout << "NO" << endl;
}

//-----------------------------
//comment from here for TOPCODER 
int32_t main(){
   IOS;
   solve();
   //find();
   #ifdef LOCAL_DEFINE
   cerr << "time elapsed " << (double)clock()/CLOCKS_PER_SEC << endl;
   #endif
}
