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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

long long  factorial(long long n) {
    if (n > 0) {
        return n * factorial(n - 1);
    } else {
        return 1;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}

//int ()



int main()
{
  int a ,b, c, d, e, X, Y;
//  long long  a, b, c, d, e, X, Y;
  //  double  c;
//  int  A[200003]={};
  long long  A[100003]={};
  long long  B[100003]={};
  int  C[100003][11]={};

  int  flg=0;

//  long long  a, b, c, d, X, Y;
  char s[101][101]={};
  char t[101]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p,q;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int len=0;
  int l=0;
  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  int ans=0;
  int max=0;
  int sum=0;
  int addr=0;
  int ptn=0;
  long long  tmp=0;
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%d", &a);
//  scanf("%lld", &A[0]);
//  A[0]=A[0]-1;
//  for(i=1;i<a;i++){
//    scanf("%lld", &A[i]);
//    if(A[i-1]<A[i]) A[i]=A[i]-1;
//    else if(A[i-1]>A[i]){
//      printf("No\n");
//      return 0;
//    }
//  }
      
  printf("%d\n", a*a*a);
 
  

    
//    printf("%s %d %d\n", s[i],C[i][0],C[i][1]);
   // printf("%d\n", C[i][1]);
/*  for(i=0;i<addr;i++){
//  scanf("%lld%lld", &B[0][i],&B[1][i]);
//    ans=ans+A[i];
    printf("%lld ", A[i]);
  }
*/
//    printf("%d ", cnt);
   

return 0;
}