Perfect Cubes

For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the "perfect cube" equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.

Input

One integer N (N <= 100).

Output

The output should be listed as shown below, one perfect cube per line, in non-decreasing order of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in non-decreasing order on the line itself. There do exist several values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first.

Sample Input

24

Sample Output

Cube = 6, Triple = (3,4,5)Cube = 12, Triple = (6,8,10)Cube = 18, Triple = (2,12,16)Cube = 18, Triple = (9,12,15)Cube = 19, Triple = (3,10,18)Cube = 20, Triple = (7,14,17)Cube = 24, Triple = (12,16,20)

//依旧暴力……
#include<stdio.h>#include<math.h>int main(){int n,s1,s2;int i,j,k,l;while(scanf("%d",&n)!=EOF){for(i=6;i<=n;i++)for(j=2;j<i;j++)for(k=j;k<i;k++)    for(l=k;l<i;l++)   if(i*i*i==j*j*j+k*k*k+l*l*l){printf("Cube = %d, Triple = (%d,%d,%d)\n",i,j,k,l);}}   return 0;}