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)

// بِسمِ اللہِ  الرَّحم?نِ  الرَّ حِیم
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define MOD 1000000007
#define ll long long
#define pb push_back
#define all(v) v.begin(),v.end()
#define INF 1e18
#define N 1001000
typedef pair<ll,ll> pll;
ll power(ll a,ll b)
{
    if(b==1)return a;
    if(b==0)return 1;
    ll temp=power(a,b/2);
    ll ans=temp*temp;
    if(b&1)
        ans*=a;
    return ans;
}
ll power(ll a,ll b,ll p)
{
    a=a%p;
    if(b==1)return a;
    if(b==0)return 1;

    ll temp=power(a,b/2,p);
    ll ans=(temp*temp)%p;
    if(b&1)
        ans=(a*ans)%p;
    return ans%p;
}
ll inv(ll a,ll m)
{
    return power(a,m-2,m);
}
//code begins
ll binary(ll a[],ll n,ll key)
{
    ll l=0,r=n-1,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(a[mid]==key)
            return mid;
        else if(a[mid]>key)
            r=mid-1;
        else
            l=mid+1;
    }
    return -1;
}
ll lowerbound(ll a[],ll n,ll key)
{
    ll l=0,r=n-1,mid,ans=-1;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(a[mid]==key)
        {
            r=mid-1;
            ans=mid;
        }
        else if(a[mid]<key)
            l=mid+1;
        else
            r=mid-1;
    }
    return ans;
}
ll upperbound(ll a[],ll n,ll key)
{
    ll l=0,r=n-1,mid,ans=-1;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(a[mid]==key)
        {
            ans=mid;
            l=mid+1;
        }
        else if(a[mid]>key)
            r=mid-1;
        else
            l=mid+1;
    }
    return (ans+1==n?-1:ans+1);
}
signed main()
{
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);

    cin.sync_with_stdio(false);
    ll t;
    t=1;
//    cin>>t;
    while(t--)
    {
        string s;cin>>s;
        if(s[2]==s[3] && s[4]==s[5])
            cout<<"Yes\n";
        else
            cout<<"No\n";
    }
    return 0;
}