Problem:
Score : 100 points

Problem Statement

A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.

Given a string S, determine whether it is coffee-like.

Constraints

S is a string of length 6 consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is coffee-like, print Yes; otherwise, print No.

Sample Input 1

sippuu

Sample Output 1

Yes

In sippuu, the 3-rd and 4-th characters are equal, and the 5-th and 6-th characters are also equal.

Sample Input 2

iphone

Sample Output 2

No

Sample Input 3

coffee

Sample Output 3

Yes

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

#include"bits/stdc++.h"
using namespace std;
typedef unsigned long long ll;


using Graph = vector<vector<int>>;//グラフ型
void dfs(const Graph& G, int v/*根*/, vector<bool>& seen) {
	seen[v] = true;  //vを訪問済にする
	for (auto next_v : G[v]) { //訪問済にしたvとつながっている場所を順番に探す
		if (seen[next_v]) continue; //もし探索済みならスルー
		dfs(G, next_v, seen); //再帰的に探索
	}
}

void bfs(const Graph& G, queue<int>& que/*発見したが訪問はしていない*/, vector<int>& dist/*開始地点からの距離*/) {
	
	while (!que.empty()) {
		int v = que.front();//queのうち先頭のものからスタート
		que.pop();//先頭を削除

		for (int nv : G[v]) {//vとつながっている点を列挙
			if (dist[nv] != -1) continue;//発見済(=距離が格納されている)であればスルー
			dist[nv] = dist[v] + 1;//vとつながっている点に距離の情報を追加
			que.push(nv);//vとつながっている点を発見済にする
		}
	}
}

int main(void) {
	string s;
	cin >> s;
	if (s[2] == s[3] && s[4] == s[5]) {
		cout << "Yes";
	}
	else {
		cout<<"No";
	}
	return 0;
}

/*
2 1  1
3 3  2
4 6  3
5 10 4

*/