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)

//rohitaas_15
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> o_set;
#define um<a,b> unordered_map<a,b>
#define ll  long long
#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define all(v) v.begin(),v.end()
#define f(a,b) for(ll i=a;i<=b;i++)
#define pb push_back
#define mp make_pair
#define pii pair<ll,ll>
#define lb(v,x) lower_bound(v.begin(),v.end(),x)
#define ub(v,x) upper_bound(v.begin(),v.end(),x)
#define show(x) cout<<x<<"\n"
#define printvector(v,n) f(0,n-1){cout<<v[i]<<" ";}
#define input(v,n) f(0,n-1){cin>>v[i];}
#define vr vector<ll>
#define check() cout<<"hi"<<"\n"
#define MOD 1000000007
#define endl "\n"
#define maxv(v) *max_element(all(v));
#define minv(v) *min_element(all(v));
#define sumv(v) accumulate(all(v),0)
#define test() ll t;cin>>t;while(t--)
ll search(vr arr, ll l, ll r, ll x){if (r >= l) {ll mid = l + (r - l) / 2;if (arr[mid] == x)return mid;if (arr[mid] > x)return search(arr, l, mid - 1, x);return search(arr, mid + 1, r, x);} return -1;}
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll lcm(ll a , ll b){return (a*b)/ gcd(a,b);}
ll power(ll x, ll y, ll p){ll res = 1;x = x % p;while (y > 0){ if (y & 1) res = (res*x) % p; y = y>>1;x = (x*x) % p; } return res; }
ll modInverse(ll n, ll p){return power(n, p-2, p);}
ll CmP(ll n, ll r, ll p=MOD){if (r==0)return 1;ll fac[n+1];fac[0] = 1;for (ll i=1 ; i<=n; i++)fac[i] = fac[i-1]*i%p; return (fac[n]* modInverse(fac[r], p) % p *modInverse(fac[n-r], p) % p) % p; }
int main()
{
	fastio();
	string s;
	cin>>s;
	if(s[2]==s[3] && s[4]==s[5])
	{
		show("Yes");
	}
	else
	{
		show("No");
	}
}