Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

rng gorilla apple

Sample Output 1

YES

They form a word chain.

Sample Input 2

yakiniku unagi sushi

Sample Output 2

NO

A and B form a word chain, but B and C do not.

Sample Input 3

a a a

Sample Output 3

YES

Sample Input 4

aaaaaaaaab aaaaaaaaaa aaaaaaaaab

Sample Output 4

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define pq priority_queue
using namespace std; typedef pair<int, int> P; typedef pair<int, P> P2; typedef pair<int, P2> P3;
constexpr long long gcd(long long a, long long b){return b ? gcd(b, a % b) : a;}
constexpr long long lcm(long long a, long long b){return a / gcd(a, b) * b;}
constexpr int INF = 1e9, MOD = INF + 7, around[] = {0, 1, 1, -1, -1, 0, -1, 1, 0, 0};
constexpr int mod_pow(long long x, long long n, const int mod){long long ret=1;while(n){if(n&1)(ret*=x)%=mod;(x*=x)%=mod;n>>=1;}return ret;}
template<int n> struct Prime{bool arr[n+1];constexpr bool operator[](int k){return arr[k];}constexpr Prime():arr(){for(int i=2;i<n;i++){arr[i]=true;for(int j=2;j*j<=i;j++){if(!(i%j))arr[i]=false;}}}};
template<int n> struct Factorial{long long arr[n+1],ary[n+1];constexpr Factorial():arr(),ary(){arr[0]=1;ary[0]=1;for(int i=0;i<n;i++){arr[i+1]=arr[i]*(i+1)%MOD;ary[i+1]=mod_pow(arr[i+1],MOD-2,MOD);}}};
constexpr Factorial<1010> fact; constexpr Prime<10> prime;
constexpr long long comb(int a, int b){if(a < b) return 0LL;if(!a or !b) return 1LL; long long pos = fact.arr[a], pot = fact.ary[a - b], por = fact.ary[b]; return pos * pot % MOD * por % MOD;}
template<int n> struct Bernoulli{long long arr[n+1];constexpr Bernoulli():arr(){arr[0]=1;for(int i=1;i<=n;i++){long long sum=0;for(int j=0;j<i;j++){(sum+=comb(i+1,j)*arr[j]%MOD)%=MOD;}arr[i]=(MOD-mod_pow(i+1,MOD-2,MOD))%MOD*sum%MOD;}}};
constexpr int vx[] = {1, 0, -1, 0}, vy[] = {0, 1, 0, -1};
constexpr long double PI = abs(acos(-1));
constexpr int sqrtN = 512, logN = 32;
constexpr long long LINF=1e18;

int main(){
	string a, b, c; cin >> a >> b >> c;
	
	if(a.back() == b[0] and b.back() == c[0]) cout << "Yes" << endl;
	else cout << "No" << endl;
	
	
	return 0;
}