stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define PI acos(-1)

typedef long long ll;

int main()

{

    char s[10];

    cin>>s;

    if(s[2]==s[3] && s[4]==s[5])

    {

        cout<<"Yes\n";



    }

    else

    {

        cout<<"No\n";

    }

    return 0;

    

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main()

{

	char S[7];

	scanf ("%s", &S);

	

	((S[2] == S[3]) && (S[4] == S[5])) ? puts ("Yes") : puts ("No");

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define ll long long

#define Arman_Sykot ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

 

using namespace std;



int main()

{

/*********************/

// Author:

        Arman_Sykot;

/*********************/

    

    string s;

    cin >> s;

    if (s[2] == s[3] && s[4] == s[5])

		cout << "Yes" << "\n";

		

	else

		cout << "No" << "\n";

    

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main()

{

	char S[7];

	scanf ("%s", &S);

	

	((S[2] == S[3]) && (S[4] == S[5])) ? puts ("Yes") : puts ("No");

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main()

{

    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    string s;

    cin >> s;

    if(s[2] == s[3] && s[4] == s[5])

        cout << "Yes";

    else

        cout << "No";

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main()

{

	char S[7];

	scanf ("%s", &S);

	

	((S[2] == S[3]) && (S[4] == S[5])) ? puts ("Yes") : puts ("No");

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include<vector>

#include<algorithm>

#include<set>

#include<map>

#include<queue>

#include<deque>

#include<algorithm>

#include<math.h>

using namespace std;



int main()

{

string s ; cin>>s;

if(s.length()<5)

{

    cout<<"No";

    return 0;

}

if(s[2]==s[3]&&s[4]==s[5])

{

    cout<<"Yes";

    return 0 ;

}

cout<<"No";

       return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	char a[6];

	scanf("%s",a);

	(a[2]==a[3]&&a[4]==a[5]) ? puts("Yes") : puts("No");



	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<string>

using namespace std;

int main()

{

	int x = 1;

	string s;

	cin>> s;

	for(int i = 0;i < s.length();i++) {

		if (s[i+2] == s[i + 3] && s[i + 4] == s[i + 5]) {

			cout << "Yes";

			break;

		}

		else {

			cout << "No";

			break;

		}

	}

	

	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	char a[6];

	scanf("%s",a);

	(a[2]==a[3]&&a[4]==a[5]) ? puts("Yes") : puts("No");



	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



int main() {

    ios::sync_with_stdio(false); cin.tie(0);

    string s;

    cin >> s;

    if (s[2] == s[3] && s[4] == s[5])

        cout << "Yes" << endl;

    else

        cout << "No" << endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	char a[6];

	scanf("%s",a);

	(a[2]==a[3]&&a[4]==a[5]) ? puts("Yes") : puts("No");



	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h> 

using namespace std; 



int main() {

     string str;

     cin >> str;

     

     if(str[2] == str[3] && str[4] == str[5])

          cout << "Yes";

     else

          cout << "No";

     

    return 0; 

} 

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

	char a[6];

	scanf("%s",a);

	(a[2]==a[3]&&a[4]==a[5]) ? puts("Yes") : puts("No");



	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;







int main()

{

    

	ios_base::sync_with_stdio(false); cin.tie(NULL);

 



	string s;

	cin>>s;



	if(s[2]==s[3] && s[4]==s[5])

		cout<<"Yes\n";

	else

		cout<<"No\n";

	

    return 0;

		

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

	char s[10];

	scanf("%s", &s);

//ex : coffee berarti f pertama = berindeks 2

					//f kedua = index 3

					//e pertama= index 4

					//e kedua = index ke 5

					//c adalah index ke 0

	if (s[2]==s[3]&& s[4]==s[5]){

		puts("Yes");

		

	}	else

		puts("No");

	

	return 0;



}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define endl  "\n"

#define ll long long

#define PI acos(-1.0)

#define test cout<<"\n****\n"

#define precise fixed(cout);cout<<setprecision(12)

#define fast  ios_base :: sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

using  namespace  std;





int main(){



    string str;

    cin>>str;



    if(str[2]==str[3] && str[4]==str[5]){

        cout<<"Yes\n";

    }

    else{

        cout<<"No\n";

    }

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

	char s[10];

	scanf("%s", &s);

//ex : coffee berarti f pertama = berindeks 2

					//f kedua = index 3

					//e pertama= index 4

					//e kedua = index ke 5

					//c adalah index ke 0

	if (s[2]==s[3]&& s[4]==s[5]){

		puts("Yes");

		

	}	else

		puts("No");

	

	return 0;



}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define ll long long

#define endl "\n"

#define mset(arr, val) memset((arr), (val), sizeof(arr))

#define For(i, init, n) for (int64_t (i) = (init); (i) < (n); (i)++)

#define ___ ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;



// const double PI = acos(-1.0), E = exp(1.0), EPS = 1e-9;

const int MOD = (int)1e9 + 7, N = (int)1e5;



int main()

{   

#ifdef MY_PREPROCESSOR

    freopen("in.txt", "r", stdin);

    freopen("out.txt", "w", stdout);

#endif

    ___

    

    string s;

    cin >> s;

    if (s[2] == s[3] && s[4] == s[5])

        cout<< "Yes";

    else

        cout<< "No";

    

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main()

{

	char s[10];

	scanf("%s", &s);

//ex : coffee berarti f pertama = berindeks 2

					//f kedua = index 3

					//e pertama= index 4

					//e kedua = index ke 5

					//c adalah index ke 0

	if (s[2]==s[3]&& s[4]==s[5]){

		puts("Yes");

		

	}	else

		puts("No");

	

	return 0;



}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

string s;

int main() {

cin>>s;

if(s[2]==s[3] && s[4]==s[5]) {

	cout<<"Yes"<<endl;

} else {

	cout<<"No"<<endl;

}

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char a[10];

	scanf("%s", &a);

	

	if(a[2]==a[3] && a[4]==a[5]){

		puts("Yes");

	}

	else

	puts("No");

	

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <string.h>

using namespace std;

string s;

int main()

{

    cin>>s;

    if (s[2]== s[3] && s[4] == s[5])cout<<"Yes"<<endl;

    else cout<<"No"<<endl;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;



int main()

{

    string s; cin >> s;

    if(s.size() < 6)

    {

        cout << "No\n";

    }

    else

    {

        if(s[2] == s[3] && s[4] == s[5]) cout << "Yes\n";

        else cout << "No\n";

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int maxn = 1e6 + 10; 

const double pi = acos(-1.0);



inline ll read()

{

    ll kk=0,f=1;

    char cc=getchar();

    while(cc<'0'||cc>'9'){if(cc=='-')f=-1;cc=getchar();}

    while(cc>='0'&&cc<='9'){kk=(kk<<1)+(kk<<3)+cc-'0';cc=getchar();}

    return kk*f;

}



char s[10];

int main(void)

{

	scanf("%s",s);

	if(s[2]==s[3]&&s[4]==s[5])printf("Yes");

	else printf("No");

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

using namespace std;



char ss[10];

int main(){

        scanf("%s", ss);

       if(ss[2]==ss[3]&&ss[4]==ss[5])

            printf("Yes\n");

       else

           printf("No\n");

       return 0;



 }


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define mod 1000000007

#define N 100005

#define ll long long

#define ull unsigned long long

#define loop1(i, j, k, step) for (int i = j; i < k; i += step)

#define loop(i, j, k) for (int i = j; i < k; i++)

#define rloop1(i, j, k, step) for (int i = j; i >= k; i -= step)

#define rloop(i, j, k) for (int i = j; i >= k; i--)

#define rep(i, k) for (int i = (0); i < (k); i++)

#define rrep(i, k) for (int i = k - 1; i >= 0; i--)



#define forall(it, l) for (auto it = l.begin(); it != l.end(); it++)

#define mp make_pair

#define pb push_back

#define pop pop_back

#define F first

#define S second

#define sz(x) (int)(x).size()

#define endl "\n"

#define Max (int)INT_MAX

#define Min (int)INT_MIN

#define MEM(a, b) memset(a, (b), sizeof(a))

#define fast_io ios_base::sync_with_stdio(0), cin.tie(NULL)

#define lin puts("");

typedef vector<int> vi;

typedef pair<int, int> pi;

typedef vector<ll> vl;

typedef pair<ll, ll> pll;

//char ch[101][101];

char c;

int arr[10000], brr[105];



void solve()

{

    string s;

    cin >> s;

    if (s.size() < 6 || s.size() > 6)

    {

        cout << "No" << endl;

        return;

    }

    if (s[2] == s[3] && s[4] == s[5])

    {

        cout << "Yes" << endl;

    }

    else

    {

        cout << "No" << endl;

    }

}

int main()

{

    auto start = clock();

    fast_io;

    //freopen("test.txt", "r", stdin);

    int t;

    t = 1;

    // cin >> t;

    while (t--)

    {

        solve();

    }

    auto stop = clock();

    cerr << "time taken:" << double(stop - start) / double(CLOCKS_PER_SEC) << " s" << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define mp make_pair

#define ff first

#define ss second

#define pb push_back

#define w(x) scanf("%d",&x);while(x--)

#define rep(i,a,b) for(i=a;i<=b;i++)

#define ios ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define pi 2*acos(0.0)

using namespace std;

typedef long long ll;

typedef pair<ll,ll> pll;

typedef vector<int> vi;



long long int mod=1e9+9;

const int N=100000;

int t,i,j,cs=0,n,a,b;

string s;



int main()

{

    cin>>s;

    if(s[2]==s[3] && s[4]==s[5])

        cout<<"Yes";

    else

        cout<<"No";

    cout<<endl;

}






// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main(){

	std::ios::sync_with_stdio(false);

	cin.tie(0);

	string s;

	cin>>s;

	if(s[2]==s[3]&&s[4]==s[5]) cout<<"Yes"<<endl;

	else cout<<"No"<<endl;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <cstring>

#include <iostream>

#include <cstdio>

#include <cstdlib>

using namespace std;

int main(){

    //freopen("data.in","r",stdin);

    string s;

    cin>>s;

    if(s[2]==s[3]&&s[4]==s[5]){

        cout<<"Yes"<<endl;

    }

    else{

        cout<<"No"<<endl;

    }

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    string s;

    cin>>s;

    char a,b,c,d;

    a=s[2];

    b=s[3];

    c=s[4];

    d=s[5];

    if(a==b && c==d)

    {

        cout<<"Yes"<<endl;

    }

    else

    {

        cout<<"No"<<endl;

    }

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main() 

{

	string s;

	cin>>s;

	if((s[2]==s[3]) && (s[4]==s[5]))

	printf("Yes\n");

	else

	printf("No\n");

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



typedef long long ll;



using namespace std;



int main()

{

    ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);



    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);



    string s;

    cin >> s;

    for(int i = 0; i<=5; i++)

    {

        if(s[2] == s[3] && s[4] == s[5])

        {

            cout << "Yes" << endl;

            break;



        }



        else

        {

            cout << "No" << endl;

            break;

        }

    }

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define io ios_base::sync_with_stdio(false); cin.tie(NULL);



#define nl "\n"

#define fr first

#define sc second

#define MP make_pair

#define pb push_back

#define VI vector<int>

#define VII vector<ll>

#define MAP map <int,int>

#define PI pair < int , int >

#define sz(v) (int)(v).size()

#define all(v) v.begin(),v.end()

#define mem(a,val) memset(a,val,sizeof(a))

#define precise(n) fixed(cout);cout<<setprecision(n)



const int N = (int)4e7;

const int MOD = (int)1e9 + 7;

const double pi = 2 * acos(0.0);

const double eps = 1e-9;



typedef long long ll;

typedef double db;



using namespace std;



int main()

{

    io;

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);



    ///Let the showdown begin





    string s;

    cin >> s;



    if((s[2]==s[3]) && (s[4]==s[5]))

        cout << "Yes" << nl;

    else

        cout << "No" << nl;



    ///Happily DONE



    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	char C[7];

		scanf("%s",C);

    		if(C[2]==C[3]&&C[4]==C[5]){

    			puts("Yes");

			}

    		else{

    			puts("No");

			}



	

 return 0;	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

int main()

{

   string s;

   cin>>s;

   if(s[2]==s[3]&&s[4]==s[5]){

    cout<<"Yes";

   }

   else{

    cout<<"No";

   }

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

int main()

{

	string s;

	cin >> s;

	if (s[2] == s[3] && s[4]==s[5])

	{

		cout << "Yes" << endl;

	}

	else

	{

		cout << "No"<<endl;

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



char c[6];



int main() {

    scanf("%s", c);

    if (c[2] == c[3] && c[4] == c[5])

        printf("Yes\n");

    else

        printf("No\n");

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;



string s;



int main(){

	cin>>s;

	if (s.length()==6){

		if (s[2]==s[3]&&s[4]==s[5]) {

			cout<<"Yes"<<endl;

			return 0;

		}

		

	}

	cout<<"No"<<endl;

	return 0; 

} 

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;

#define ufor(i,l,r) for (int i=l;i<=r;i++)

#define MAXN 20000500



int main()

{

	char s[100000];

	scanf("%s",s);

	bool flag=(s[2]==s[3] && s[4] == s[5]);

	printf((flag)? "Yes":"No");

		

	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <string>

using namespace std;

int main(int argc, const char * argv[]) {

    ios::sync_with_stdio(false);

    string s;

    cin>>s;

    if (s.size()!=6) {

        

    }else{

        if (s[2]==s[3]&&s[4]==s[5]) {

            cout<<"Yes";

        }else{

            cout<<"No";

        }

    }

    

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    string s;

    cin>>s;



    if(s[2] == s[3] && s[4] == s[5])

        cout<<"Yes";

    else

        cout<<"No";

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#include<string.h>



using namespace std;



int main()

{

    char line[7];

    for(int i=0;i<6;i++)

    {

        cin>>line[i];

        if(i==3)

        {

            if(line[i]!=line[i-1])

            {

               cout<<"No";

               return 0;

            }

        }

        else if(i==5)

        {

            if(line[i]!=line[i-1])

            {

               cout<<"No";

               return 0;

            }

        }

    }

    cout<<"Yes";

    return 0;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <stdlib.h>

#include <string.h>



using namespace std;



int main()

{

    char str[80];

    cin>>str;

    if((str[2]==str[3]) && (str[4]==str[5]))

       cout<<"Yes";

    else

        cout<<"No";



    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

int main()

{

    string s;

    cin >> s;

    if(s[2]==s[3] && s[4]==s[5])

    {

        cout << "Yes" << endl;

    }

    else cout << "No" << endl;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

int main()

{

    char S[7];

    cin >> S;

    if(S[2]== S[3]) {

        if(S[4]==S[5]){

       cout << "Yes" << endl;

        }

        else cout << "No" << endl;

     }

    else cout << "No" << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main () {

	char c1, c2, c3, c4, c5, c6;

	scanf("%c%c%c%c%c%c", &c1, &c2, &c3, &c4, &c5, &c6);

	

	c3 == c4 && c5 == c6 ? puts("Yes") : puts("No");

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define Tayeb ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)

typedef long long ll;

using namespace std;

int main()

{

    Tayeb;

    string s;

    cin >> s;



    if(s[2]==s[3]&&s[4]==s[5])

    cout << "Yes\n";

    else

    cout << "No\n";



    return 0;



}


// Optimized solution to the same problem (faster):
#include<stdio.h>





 int main(){

 char S[7];

 scanf("%s", &S);

 if(S[2]==S[3] && S[4] == S[5]){

 	puts("Yes");

 }

 else{

 	puts("No");

 }

 return 0;

	}


// Fastest known solution for this problem: