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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
  
using namespace std;
using namespace __gnu_pbds;
 
#define ll long long
#define uint unsigned int
#define ull  unsigned long long
#define pb push_back
#define mk make_pair
#define ins insert
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define X first
#define Y second
#define umap unordered_map
#define speed() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define setvalue(d,s,e,n) for(int qwe = s; qwe < e; ++qwe) d[qwe] = n
#define mset multiset
#define pqueue priority_queue
   
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
        
const int  N = 2e5 + 314;
const int INF = 1e9;
const double PI = acos(-1);
const int MOD = 1e9 + 7;
const double eps = 1e-9;
 
void solve(){
    string a, b, c;
    cin >> a >> b >> c;
    if(a[a.size() - 1] == b[0] && b[b.size() - 1] == c[0]) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
 
int main(){
    speed();
    int t = 1;
    // cin >> t;
    while(t--)solve();
}