Problem:
Score : 100 points

Problem Statement

Takahashi is going to set a 3-character password.

How many possible passwords are there if each of its characters must be a digit between 1 and N (inclusive)?

Constraints

1 \leq N \leq 9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of possible passwords.

Sample Input 1

2

Sample Output 1

8

There are eight possible passwords: 111, 112, 121, 122, 211, 212, 221, and 222.

Sample Input 2

1

Sample Output 2

1

There is only one possible password if you can only use one kind of character.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace::std;
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/numeric/interval.hpp>
#include <boost/numeric/interval/io.hpp>
#include <boost/intrusive/rbtree.hpp>
#include <boost/geometry.hpp>
#include <boost/geometry/geometries/linestring.hpp>
#include <boost/geometry/geometries/polygon.hpp>
#include <boost/geometry/geometries/point_xy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds;
using namespace::std;
namespace mp = boost::multiprecision;
typedef mp::number<mp::cpp_dec_float<0>> cfloat;
typedef mp::cpp_int cint;
typedef long long lint;
typedef long long ll;
typedef long double ldouble;
typedef vector<lint> vec;
typedef vector<vector<lint>> mat;
typedef vector<vector<vector<lint>>> mat3;
typedef vector<double> dvec;
typedef vector<vector<double>> dmat;
typedef vector<vector<vector<double>>> dmat3;
typedef vector<string> svec;
typedef vector<vector<string>> smat;
typedef vector<vector<vector<string>>> smat3;
typedef vector<pair<lint,lint>> pvec;
typedef vector<vector<pair<lint,lint>>> pmat;
typedef vector<vector<vector<pair<lint,lint>>>> pmat3;
#define rep(i, n) for(lint i = 0; i < (lint)(n); i++)
#define irep(i) for(lint i = 0;; i++)
#define irep1(i) for(lint i = 1;; i++)
#define irep2(i) for(lint i = 2;; i++)
#define rrep(i, n) for(lint i = (lint)(n-1); i >-1; i--)
#define rrepi(i,a,b) for(lint i = (lint)(b-1); i >a-1; i--)
#define repi(i,a,b) for(lint i=lint(a);i<lint(b);i++)
#define rep2(i,a,b,c) for(lint i=lint(a);i>lint(b);i+=c)
#define all(x) (x).begin(),(x).end()
#define PI 3.141592653589793
#define dist(x1,y1,x2,y2) (pow(pow(x2-x1,2)+pow(y2-y1,2),0.5))
#define output(v) do{bool f=0;for(auto i:v){cout<<(f?" ":"");if(i>INF/2)cout<<"INF";else cout<<i;f=1;}cout<<"\n";}while(0)
#define output2(v) for(auto j:v)output(j);
#define input(a,n) lint n;cin>>n;vector<lint>a(n);rep(i,n)cin>>a[i];
#define SUM(v) accumulate(all(v),0LL)
#define INF (1LL<<60)
#define IINF (1<<30)
#define EPS (1e-10)
#define LINF 9223372036854775807
#define MOD 1000000007
#define endl "\n"
template<typename T=lint>T in(){return *istream_iterator<T>(cin);}
inline lint gcd(lint a,lint b){return b?gcd(b,a%b):a;}
inline lint lcm(lint a,lint b){return a*b/gcd(a,b);}
inline bool chmin(auto& s,const auto& t){bool res=s>t;s=min(s,t);return res;}
inline bool chmax(auto& s,const auto& t){bool res=s<t;s=max(s,t);return res;}
vector<lint> dx={-1,1,0,0,1,1,-1,-1};
vector<lint> dy={0,0,-1,1,1,-1,1,-1};

int main(){
    lint n;
    cin>>n;
    cout<<(lint)pow(n,3);
}