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 <iostream>
#include <bitset>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
#include <regex>
using namespace std;
using ll = long long int;
using dd = long double;
const ll MOD = 1e9 + 7;

vector<string> dfs(string s, ll depth, vector<char> symbol)
{
    vector<string> ret;
    if (depth == 0)
    {
        ret.push_back(s);
        return ret;
    }

    for (ll i = 0; i < symbol.size(); i++)
    {
        auto vec = dfs(s + symbol[i], depth - 1, symbol);
        for (auto &&e : vec)
        {
            ret.push_back(e);
        }
    }
    return ret;
}

ll gcd(ll a, ll b)
{
    if (a % b == 0)
    {
        return b;
    }
    return gcd(b, a % b);
}

int main()
{
    string a, b, c;
    cin >> a >> b >> c;

    bool ok = (a.back() == b.front()) && b.back() == c.front());
    cout << (ok ? "YES" : "NO") << endl;
}