## CUBEFR - Cube Free Numbers

A cube free number is a number who’s none of the divisor is a cube number (A cube number is a cube of a integer like 8 (2 * 2 * 2) , 27 (3 * 3 * 3) ). So cube free numbers are 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18 etc (we will consider 1 as cube free). 8, 16, 24, 27, 32 etc are not cube free number. So the position of 1 among the cube free numbers is 1, position of 2 is 2, 3 is 3 and position of 10 is 9. Given a positive number you have to say if its a cube free number and if yes then tell its position among cube free numbers.

### Input

First line of the test case will be the number of test case T (1 <= T <= 100000) . Then T lines follows. On each line you will find a integer number n (1 <= n <= 1000000).

### Output

For each input line, print a line containing “Case I: ”, where I is the test case number. Then if it is not a cube free number then print “Not Cube Free”. Otherwise print its position among the cube free numbers.

### Example

```Sample Input:
10
1
2
3
4
5
6
7
8
9
10

Sample Output:
Case 1: 1
Case 2: 2
Case 3: 3
Case 4: 4
Case 5: 5
Case 6: 6
Case 7: 7
Case 8: Not Cube Free
Case 9: 8
Case 10: 9```

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;

#define MOD (ll)1000000007
#define pb 	push_back
#define EPS 1e-9
#define FOR(i, n)	for(int i = 0;i < n; i++)
#define pi(a)   printf("%d\n", a)
#define all(c)  c.begin(), c.end()
#define tr(container, it)   for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}

int cube[1000005];
vector<int> res;
void precal(){
for(int i = 2;i <= 100; i++){
cube[i*i*i] = 1;
}
for(int i = 8;i <= 1000000; i++){
if(cube[i]){
for(int j = 2*i; j <= 1000000; j+=i){
cube[j] = 1;
}
}
}
for(int i = 1;i <= 1000000; i++){
if(!cube[i])
res.pb(i);
}
}

int main(){
int t;
si(t);
precal();
int kase = 1;
while(t--){
int n;
si(n);
if(cube[n]){
printf("Case %d: Not Cube Free\n", kase);
}else{
//find pos of n in res
int pos = lower_bound(res.begin(), res.end(), n)-res.begin()+1;
printf("Case %d: %d\n", kase, pos);
}
kase++;
}
return 0;
}``````
Copy The Code &

Input

cmd
10
1
2
3
4
5
6
7
8
9
10

Output

cmd
Case 1: 1
Case 2: 2
Case 3: 3
Case 4: 4
Case 5: 5
Case 6: 6
Case 7: 7
Case 8: Not Cube Free
Case 9: 8
Case 10: 9