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<stdio.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
#include<functional>
#include<limits.h>
#include<stdlib.h>
#include<string>
#include<unordered_map>
using namespace std;
 
#define intmax INT_MAX
#define lmax LONG_MAX
#define uintmax UINT_MAX
#define ulmax ULONG_MAX
#define llmax LLONG_MAX
#define ll long long
#define rep(i,a,N) for((i)=(a);(i)<(N);(i)++)
#define rrp(i,N,a) for((i)=(N)-1;(i)>=(a);(i)--)
#define llfor ll i,j,k
#define sc(a) cin>>a
#define pr(a) cout<<a<<endl
#define pY puts("YES")
#define pN puts("NO")
#define py puts("Yes")
#define pn puts("No")
#define pnn printf("\n")
#define sort(a) sort(a.begin(),a.end())
#define push(a,b) (a).push_back(b)
#define llvec vector<vector<ll>>
#define charvec vector<vector<char>>
#define sizeoof(a,b) (a,vector<ll>(b))
#define llpvec vector<pair<ll,ll>>
/*繰り上げ除算*/ll cei(ll x,ll y){ll ans=x/y;if(x%y!=0)ans++;return ans;}
/*最大公約数*/ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*最小公倍数*/ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
/*n乗*/ll llpow(ll x,ll n){ll i,ans=1;rep(i,0,n)ans*=x;return ans;}
/*階乗*/ll fact(ll x){ll i,ans=1;rep(i,0,x)ans*=(x-i);return ans;}
/*nCr*/ll ncr(ll n,ll r){return fact(n)/fact(r)/fact(n-r);}
/*nPr*/ll npr(ll n,ll r){return fact(n)/fact(n-r);}
/*primejudge*/bool prime(ll a){if(a<=1)return false;ll i;for(i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}
ll ans=0;llfor;///////////////////////////////////////////////////////////
int main(){
  string S,T,U;
  cin>>S>>T>>U;
  int A=S.size();
  int B=T.size();
  string S="NO";
  if(S[A-1]==T[0]&&T[B-1]==U[0]) S="YES";
  cout<<S;
return 0;}