Description

Your task in this problem is to determine the number of divisors of

C_n^k. Just for fun -- or do you need any special reason for such a useful computation?

Input

The input consists of several instances. Each instance consists of a single line containing two integers n and k (0 ≤ k ≤ n ≤ 431), separated by a single space.

Output

For each instance, output a line containing exactly one integer -- the number of distinct divisors of

C_n^k. For the input instances, this number does not exceed 2

⁶³ - 1.

Sample Input

5 16 310 4

Sample Output

2616

分析：一个数的因子个数求法  将数写成 n=p1^a*p2^b..pn^c;  也就是质因素乘积的形式 那么因子的个数就是 sum=(a+1)*(b+1)*...*(c+1);

code：

#include
      
        #include
       
         int p[90]; int v[432]; int pn; int e[432][90]; void pri() {
    int i,j;     pn=0;     memset(v,0,sizeof(v)); for(i=2;i<=431;i++)     {
    if(v[i]==0)         {
                p[pn++]=i; for(j=i;j<=431;j+=i)             v[j]=1;         }     } } void fac() {
    int i,j,k; for(i=2;i<=431;i++)     {
            k=i; for(j=0;j
        
         1&&k%p[j]==0)                 { e[i][j]++; k/=p[j];}     } for(i=3;i<=431;i++) for(j=0;j