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
/*input
async layered coding
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#ifdef leowang
#define debug(...) do{\
	fprintf(stderr,"%s - %d : (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
	_DO(__VA_ARGS__);\
}while(0)
template<typename I> void _DO(I&&x){cerr<<x<<endl;}
template<typename I,typename...T> void _DO(I&&x,T&&...tail){cerr<<x<<", ";_DO(tail...);}
#else
#define debug(...)
#endif
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
	out<<'('<<P.F<<','<<P.S<<')';
	return out;
}
//}}}
const ll maxn=300005;
const ll maxlg=__lg(maxn)+2;
const ll INF64=8000000000000000000LL;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const double PI=acos(-1);
//const ll p=880301;
//const ll P=31;

ll mypow(ll a,ll b){
	ll res=1LL;
	while(b){
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		
		b>>=1;
	}
	return res;
}
int main()
{
	IOS;
	string a,b,c;
	cin>>a>>b>>C;
	if(a.back()==b[0]&&b.back()==c[0]){
		cout<<"YES\n";
	}
	else cout<<"NO"
	return 0;
}
