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)

#include <bits/stdc++.h>

using namespace std;

struct vw_pair {
    int v, w;
    vw_pair(int v,int w) : v(v), w(w) {};
};

typedef pair<int, int> ii;
typedef vector <ii> vii;
typedef vector<int> vi;
typedef vector <string> vs;
typedef vector <vi> vvi;
typedef vector <vector<vw_pair>> adj_list;
typedef vector <bool> vb;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef priority_queue<int> max_heap_int;
typedef priority_queue<int, vi, greater<>> min_heap_int;

#define isOn(S, j) (S & (1<<j))
#define setBit(S, j) (S |= (1<<j))
#define clearBit(S, j) (S &= ~(1<<j))
#define toggleBit(S, j) (S ^= (1<<j))
#define lowBit(S) (S & (-S))
#define setAll(S, n) (S = (1<<n)-1)
#define modulo(S, N) ((S) & (N-1))   // returns S % N, where N is a power of 2
#define isPowerOfTwo(S) (!(S & (S-1)))
#define nearestPowerOfTwo(S) ((int)pow(2.0, (int)((log((double)S) / log(2.0)) + 0.5)))
#define turnOffLastBit(S) ((S) & (S-1))
#define turnOnLastZero(S) ((S) | (S+1))
#define turnOffLastConsecutiveBits(S) ((S) & (S+1))
#define turnOnLastConsecutiveZeroes(S) ((S) | (S-1))
#define in_range(lo, hi, x) ((x >= lo && x <= hi))

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;
    cout << n*n*n << "\n";
    
    return 0;
}