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本文介绍了一种计算特定形式五元立方方程解数量的方法。通过预先计算并存储部分解的情况，有效地减少了整体计算量。输入为五个整数系数，输出为满足条件的解的数量。

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Consider equations having the following form:
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.

Determine how many solutions satisfy the given equation.

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

#include<stdio.h>
#include<math.h>
#include<string.h>
short jie[25000000];
int main()
{
    int a1,a2,a3,a4,a5,i,j,t,k,sum;
    scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
    memset(jie,0,sizeof(jie));
  for(i=-50;i<=50;i++)
    {
        for(j=-50;j<=50;j++)
        {
            if(i!=0&&j!=0)
            {
            t=i*i*i*a1+j*j*j*a2;
            if(t>0)
                jie[t]++;
            else
            {
                t=t+25000000;
                jie[t]++;
            }
            }
        }
    }
    sum=0;
    for(i=-50;i<=50;i++)
    {
        for(j=-50;j<=50;j++)
        {
            for(k=-50;k<=50;k++)
            {
                if(i!=0&&j!=0&&k!=0)
                {
                   t=a3*i*i*i+a4*j*j*j+a5*k*k*k;
                   t=-t;
                   if(t>0&&t<25000000)
                        sum=sum+jie[t];
                   if(t<=0&&t>-25000000)
                    {
                        t=t+25000000;
                          sum=sum+jie[t];
                    }

               }
            }
        }
    }
    printf("%d\n",sum);
    return 0;