stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.2510425656294215x):
#include <stdio.h>

 

int main ()

{

    int n;

    scanf ("%d", &n);

    printf ("%d\n", n*n*n);

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.270375595763333x):
#include <stdio.h>

 

int main ()

{

    int n;

    scanf ("%d", &n);

    printf ("%d\n", n*n*n);

    return 0;

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.3232620787629585x):
#include <stdio.h>

int main (){

	int character;

	int n;

	

	scanf("%d", &character);

	printf("%d\n", n = character*character*character);

	

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.2510425656294215x):
#include <stdio.h>

int main (){

	int character;

	int n;

	

	scanf("%d", &character);

	printf("%d\n", n = character*character*character);

	

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.270375595763333x):
#include <stdio.h>

int main (){

	int character;

	int n;

	

	scanf("%d", &character);

	printf("%d\n", n = character*character*character);

	

	

	return 0;

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.328714601507691x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	int kode = n*n*n;

	printf("%d", kode);

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.256421115210304x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	int kode = n*n*n;

	printf("%d", kode);

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.275773947823404x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	int kode = n*n*n;

	printf("%d", kode);

	return 0;

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.3232620787629585x):
#include <stdio.h>



int main() {

	

	int number;

	

	scanf("%d", &number);

	printf("%d\n", number*number*number);

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.2510425656294215x):
#include <stdio.h>



int main() {

	

	int number;

	

	scanf("%d", &number);

	printf("%d\n", number*number*number);

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.270375595763333x):
#include <stdio.h>



int main() {

	

	int number;

	

	scanf("%d", &number);

	printf("%d\n", number*number*number);

	

	return 0;

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.3232620787629585x):
#include <stdio.h>

int main (){

	int n;

	scanf("%d",&n);

	int password = n*n*n;

	printf("%d\n",password);

	

	

	

	

	return 0;	

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.2510425656294215x):
#include <stdio.h>

int main (){

	int n;

	scanf("%d",&n);

	int password = n*n*n;

	printf("%d\n",password);

	

	

	

	

	return 0;	

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.270375595763333x):
#include <stdio.h>

int main (){

	int n;

	scanf("%d",&n);

	int password = n*n*n;

	printf("%d\n",password);

	

	

	

	

	return 0;	

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.328714601507691x):
#include <stdio.h>



int main()

{

    int var;

    scanf("%d",&var);

    printf("%d",var*var*var);

   

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.256421115210304x):
#include <stdio.h>



int main()

{

    int var;

    scanf("%d",&var);

    printf("%d",var*var*var);

   

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.275773947823404x):
#include <stdio.h>



int main()

{

    int var;

    scanf("%d",&var);

    printf("%d",var*var*var);

   

    return 0;

}
// Slower C++ solution:
/*ANKON GHOSH

  KUET,ECE 2K16*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define inf 1e18

#define MAX 4003

#define mod 1000000007

#define sf(a) scanf("%lld",&a)

#define sff(a,b) scanf("%lld %lld",&a,&b)

#define sfff(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define Case(i) cout<<"Case #"<<i<<": "

#define pf(a) printf("%lld\n",a)

#define pfsn printf("NO\n")

#define pfs  printf("YES\n")

#define pff(a,b) printf("%lld %lld\n",a,b)

#define pfff(a,b,c) printf("%lld %lld %lld\n",a,b,c)

#define pffff(a,b,c,d) printf("%lld %lld %lld %lld\n",a,b,c,d)

#define FOR(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define LOOP(x,y) for(ll r=x;r<=y;r++)

#define RFOR(i,n) for(ll i=n-1;i>=0;i--)

#define RFOR1(i,n) for(ll i=n;i>0;i--)

#define RLOOP(x,y) for(ll i=y;i>=x;i--)

#define m_p(a,b) make_pair(a,b)

#define p_b(a) push_back(a)

#define ff first

#define ss second

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

map<ll,ll>mp1,mp;

vector<pair<ll,pair<ll,ll> > >vpp,vpp1;

vector<pair<ll,ll> >vp,fin;

vector<ll>v1,v2,dummy;

map<pair<ll,ll>,bool>mbb;

map<ll,bool>mb;

ll n,k,a,b,c,d,ta,sum=0,cnt=0,m,l,req,sz,maxi;

bool sortbysec(const pair<pair<ll,ll>,ll> &a,const pair<pair<ll,ll>,ll> &b)

{

    return (a.second < b.second);

}

vector<ll>power;

int main()

{

    cin>>n;

    cout<<powl(n,3LL);

    return 0;

}

/*

5

4

3

2

3

4

4

3

3

4

2

3

*/


// Faster C++ solution for the same problem (measured speedup: 5.3232620787629585x):
#include <stdio.h>

int main() {

	int n;

	scanf("%d",&n);

	int total= n*n*n;

	printf("%d\n",total);

return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define ll long long int

#define co cout

#define ci cin

#define sf1(n) scanf("%lld",&n);

#define sf2(n,m) scanf("%lld %lld",&n,&m);

#define sf3(n,m,p) scanf("%lld %lld %lld",&n,&m,&p);

#define en cout<<endl;

#define pb push_back

#define coy cout<< "YES"<<endl;

#define con cout<< "NO"<<endl;

#define For(a,b) for(ll a=0;a<b;a++)

#define br break;

#define i64 long long

i64 M=10000000;

ll b[805],ans[805];

map< ll,pair<ll,ll> >mp;

//vector<ll>a;

int main()

{

    ll t,n,g,k,m,mx=0,mn=1e9,c,res,c1,x,y,c2;

    sf1(t)

    co<<t*t*t;

}




// Faster C++ solution for the same problem (measured speedup: 5.2510425656294215x):
#include <stdio.h>

int main() {

	int n;

	scanf("%d",&n);

	int total= n*n*n;

	printf("%d\n",total);

return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

int n;



cin>>n;



long long int a = pow(n,3);

cout<<a<<endl;

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.270375595763333x):
#include <stdio.h>

int main() {

	int n;

	scanf("%d",&n);

	int total= n*n*n;

	printf("%d\n",total);

return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#include <math.h>

using namespace std;



int n;

int main () {

cin>>n;

cout<<pow(n,3)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.345943631817677x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

int main(){

int n;

cin >> n;

cout << n*n*n << endl;

return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.270865050861552x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include<cmath>



using namespace std;



int main()

{

    int n;

    cin>>n;

    cout<<pow(n,3);



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.3365798837719005x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;







int main()

{

  int N;

  cin>>N;

int x =N*N*N;

cout<<x;



		return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.261629022032166x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cmath>

using namespace std;



int main()

{

    int n;

    cin >> n;

    int x=pow(n,3);

    cout << x;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.268518168871263x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include<cmath>

using namespace std;



int main()

{

    int b;

    cin>>b ;

    if(b==1)

        cout<<b;

    else cout<<pow(b,3);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.336543882786088x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std; 

int main(){

int N;

cin >> N; 

cout<<N*N*N;

return 0; }

// Faster C++ solution for the same problem (measured speedup: 5.261629022032166x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

int main(){

    int n;

    cin>>n;

    cout<<n*n*n;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.261629022032166x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

	ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    int n;

    cin>>n;

    cout<<n*n*n;

}

// Faster C++ solution for the same problem (measured speedup: 5.337633680186476x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <iostream>

//#include <stdlib.h>

//#include <string.h>



using namespace std;



int main()

{

    int n;

    cin>>n;

    cout<<n*n*n;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.261629022032166x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	printf("%d", n * n * n);

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#include <math.h>

using namespace std;



int n;

int main () {

cin>>n;

cout<<pow(n,3)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.340473479740795x):
#include <stdio.h>



int main(){

    int N;

    scanf("%d", &N);

    

    printf("%d\n", N*N*N);



    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

int main(){

int n;

cin >> n;

cout << n*n*n << endl;

return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.265471721752481x):
#include <stdio.h>



int main(){

    int N;

    scanf("%d", &N);

    

    printf("%d\n", N*N*N);



    return 0;

}
// Slower C++ solution:
#include <iostream>

#include<cmath>



using namespace std;



int main()

{

    int n;

    cin>>n;

    cout<<pow(n,3);



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.331119313001772x):
#include <stdio.h>



int main(){

    int N;

    scanf("%d", &N);

    

    printf("%d\n", N*N*N);



    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;







int main()

{

  int N;

  cin>>N;

int x =N*N*N;

cout<<x;



		return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.256245143543185x):
#include <stdio.h>



int main(){

    int N;

    scanf("%d", &N);

    

    printf("%d\n", N*N*N);



    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cmath>

using namespace std;



int main()

{

    int n;

    cin >> n;

    int x=pow(n,3);

    cout << x;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.263127241171985x):
#include <stdio.h>



int main(){

    int N;

    scanf("%d", &N);

    

    printf("%d\n", N*N*N);



    return 0;

}