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<algorithm>//everybody dies in their nightmares
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<map>
#include<queue>
#include<deque>
#include<string>
#include<vector>
#include<sstream>
#include<cmath>
#include<unordered_set>
#include<unordered_map>
#define f(i,a,b) for(register int i=a;i<=b;++i)
#define ff(i,a,b) for(register int i=a;i>=b;--i)
#define debug(x) cerr << #x << " : " << x << " " << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const ll mod = 1e18 + 7;
const double tiaohe = 0.57721566490153286060651209;
ll oula(ll x) { ll res = x;f(i, 2, x / i) { if (x % i == 0) { res = res / i * (i - 1);while (x % i == 0) x /= i; } }if (x > 1) res = res / x * (x - 1);return res; }
ll quickmod(ll a, ll n, ll m) { ll s = 1;while (n) { if (n & 1) { s = s * a % m; }a = (a*a) % m;n = n / 2; }return s; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
void ex_gcd(ll a, ll b, ll &x, ll &y, ll &d) { if (!b) { d = a, x = 1, y = 0; } else { ex_gcd(b, a % b, y, x, d);y -= x * (a / b); } }
ll inv(ll t, ll p) { ll d, x, y;ex_gcd(t, p, x, y, d);return d == 1 ? (x % p + p) % p : -1; }
bool isPrime(ll x) { if (x == 2)return true;if (x % 2 == 0)return false;for (ll i = 2;i*i <= x;i++) if (x % i == 0)return false; return true; }
//double a = log(n) +tiaohe + 1.0 / (2 * n);
double eqa = (1 + sqrt(5.0)) / 2.0;
const int N = 2e5 + 5;
int t;

int main()
{
	string a;
	while (cin >> a)
	{
		int tt = a.size();
		if (tt == 6 && a[2] == a[3] && a[4] == a[5])puts("Yes");
		else puts("No");
	}
	return 0;
}