// Original C++ solution (slower):
/**

DufauX-XuafuD

**/

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<long long, long long> pll;

#define ff first

#define ss second

#define mp make_pair

#define pb push_back

#define all(x) (x).begin(), (x).end()

#define FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

#define reset(x) memset(x,-1,sizeof(x))

#define CheckBit(a,b) (a&(1ll<<b))

#define SetBit(a,b) a=(a|(1ll<<b))

#define UnSetBit(a,b) a=(a&(~(1ll<<b)))

#define maxx 10000006

#define PI 2*acos(0.0)

const long long INF = 2000000000000000000LL;	// 2e18

const int inf = 0x3f3f3f3f;	// 1061109567

const long double EPS = 1e-9;

//bitset<maxx/2>vis;

//bitset<maxx>pr;

//int mobi[maxx+10];

//vector<ll>prime;

//void sieve(){

//    ll x=maxx/2, y=sqrt(maxx)/2;

//    for(ll i=1;i<=y;i++){

//        if(vis[i]==0){

//            for(ll j=(i*(i+1)*2);j<x;j+=(2*i)+1)

//                vis[j]=1;

//        }

//    }

//    prime.push_back(2);

//    for(ll i=3;i<maxx;i+=2)

//        if(vis[i/2]==0)

//        prime.push_back(i);

//}

//void generate_mobius()

//{

//    for(int i=0;i<maxx;i++)mobi[i]=1;

//    for(int i=2;i<maxx;i++){

//        if(!pr[i]){

//            mobi[i]*=-1;

//            for(int j=i+i;j<maxx;j+=i){

//                pr[j]=1;

//                mobi[j]*=(j%(i*i)?-1:0);

//            }

//        }

//    }

//}

//bool isprime(ll a)

//{

//    if(a==1)return false;

//    if(a==2)return true;

//    if(a%2==0)return false;

//    for(ll i=0;i<prime.size() && prime[i]*prime[i]<=a;i++)

//        if(a%prime[i]==0)return false;

//    return true;

//}

/* Calculate ( a*b ) %c */

inline ll mulmod(ll a, ll b, ll c) {

    ll ret = 0 , cur = a % c ;

    while(b) { if(b%2) ret=(ret+cur)%c; cur=(cur<<1)%c; b=b>>1; }

    return ret%c;

}

//Calculate (b^p)%m || modular inverse: b^-1 = bigmod(b, m-2, m)

inline ll bigmod(ll b, ll p, ll m){

    ll ret = 1, cur = b % m;

    while(p) { if(p&1) ret=(ret*cur)%m; cur=(cur*cur)%m; p=p>>1; }

    return ret;

}

inline ll modulo(ll a,ll b,ll c) { //Same as big mod, use for numbers over 1e10

    ll ret = 1 , cur = a%c ;

    while(b) { if(b%2) ret=mulmod(ret,cur,c); cur=mulmod(cur,cur,c); b=b>>1; }

    return ret%c ;

}

inline ll power(ll b, ll p){

    ll ret = 1, cur = b;

    while(p) { if(p&1) ret=(ret*cur); cur=(cur*cur); p=p>>1; }

    return ret;

}

ll lcm(ll a,ll b)

{

    return (a/__gcd(a,b))*b;

}

ll gcd(ll a,ll b)

{

    return __gcd(a,b);

}

void Dufaux()

{

    string a,b,c;

    cin>>a>>b>>c;

    if(a[a.size()-1]==b[0] && b[b.size()-1]==c[0]){

        cout<<"YES"<<endl;

    }

    else{

        cout<<"NO"<<endl;

    }

}

int main()

{

    FastIO;

	///sieve();

	///generate_mobius();

	int test=1;

	while(test--){

        Dufaux();

	}

	return 0;

}










// Optimized solution to the same problem (faster):
#include<stdio.h>

 

int main(){

  char a[100];

  scanf("%[^\n]", a);

	int ctr = 0;

	

  for(int i= 0; a[i] != '\0'; i++){

    if(a[i] == ' '){

		if(a[i - 1] == a[i + 1]){

			ctr++;

		}

    }

  }

  if(ctr == 2){

  	printf("YES\n");

  }

  else{

  	printf("NO\n");

  }

  return 0;

}

// Fastest known solution for this problem: