A positive integer may be expressed as a sum of different prime numbers (primes), in one way or another. Given two positive integers n and k, you should count the number of ways to express n as a sum of k different primes. Here, two ways are considered to be the same if they sum up the same set of the primes. For example, 8 can be expressed as 3 + 5 and 5 + 3 but the are not distinguished.

When n and k are 24 and 3 respectively, the answer is two because there are two sets {2, 3, 19} and {2, 5, 17} whose sums are equal to 24. There are not other sets of three primes that sum up to 24. For n = 24 and k = 2, the answer is three, because there are three sets {5, 19}, {7, 17} and {11, 13}. For n = 2 and k = 1, the answer is one, because there is only one set {2} whose sum is 2. For n = 1 and k = 1, the answer is zero. As 1 is not a prime, you shouldn’t count {1}. For n = 4 and k = 2, the answer is zero, because there are no sets of two different primes whose sums are 4.

Your job is to write a program that reports the number of such ways for the given n and k.

The input is a sequence of datasets followed by a line containing two zeros separated by a space. A dataset is a line containing two positive integers n and k separated by a space. You may assume that n ≤ 1120 and k ≤ 14.

The output should be composed of lines, each corresponding to an input dataset. An output line should contain one non-negative integer indicating the number of the ways for n and k specified in the corresponding dataset. You may assume that it is less than 2³¹.

 #include<iostream>

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<cstdlib>

 #include<iomanip>

 #include<cmath>

 #include<vector>

 #include<queue>

 #include<stack>

 using namespace std;

 #define PI 3.141592653589792128462643383279502

 #define N 1200

 int prim[N]={,},f[][N],t;

 int prime(){

     int t=,i,j,flag;

     for(i=;i<N;i+=){

         for(j=,flag=;prim[j]*prim[j]<=i;j++)

             if(i%prim[j]==) flag=;

         if(flag){

             prim[t++]=i;

         }

     }

     return t-;

 }

 void s(){

      for(int i=;i<=t;i++){

                 for(int j=;j>=;j--){

                     for(int p=;p>=prim[i];p--)

                         f[j][p]+=f[j-][p-prim[i]];

                 }

             }

 }

 int main(){

     //#ifdef CDZSC_June

     //freopen("in.txt","r",stdin);

     //#endif

     //std::ios::sync_with_stdio(false);

     t=prime();

     int k,n;

     while(scanf("%d%d",&n,&k)){

         memset(f,,sizeof(f));

         f[][]=;

         if(k==&&n==) break;

         s();

          cout<<f[k][n]<<endl;

     }

     return ;

 }