/*
  file : /main.comp.cpp
  date : 2015-10-23

 */

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<cassert>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;
typedef ll int__;
#define rep(i,j) for(int__ i=0;i<(int__)(j);i++)
#define repeat(i,j,k) for(int__ i=(j);i<(int__)(k);i++)
#define all(v) v.begin(),v.end()

#define N_MAX 1130
#define K_MAX 15
int DP[N_MAX][K_MAX];//n k


void init(){
    vector<bool> is_prime(N_MAX,true);
    is_prime[0]=false;
    is_prime[1]=false;
    for(int i=2;i<N_MAX;i++){
        if(not is_prime[i])continue;
        for(int j =i*2;j<N_MAX;j+=i){
            is_prime[j]=false;
        }
    }

    rep(i,N_MAX){
        rep(j,K_MAX){
            DP[i][j]=0;
        }
    }

    // rep(p,N_MAX){
    //     if(is_prime[p])DP[p][1]=1;
    // }
    DP[0][0]=1;
    rep(p,N_MAX){
        if(is_prime[p]){
            for(int k=K_MAX-1;k>=0;k--){
                rep(n,N_MAX){
                    if(n-p>=0 and k-1>=0)DP[n][k] += DP[n-p][k-1];
                    //printf("%lld %d %d\n",n,k,DP[n][k]);
                }
            }
        }


    }
}


bool solve(){
    int n,k;
    cin>>n>>k;
    if(n==0 && k==0)return false;
    cout << DP[n][k] <<endl;    
    return true;
}

int main()
{
    init();

    // rep(n,25){
    //     cout<<n<<"  ";
    //     rep(k,3){
    //         cout <<DP[n][k] <<" ";
    //     }
    //     cout <<endl;
    // }
    
    while(solve());
    return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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 they 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 no 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 diffrent 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.



Input

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.

Output

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 ways for n and k specified in the corresponding dataset. You may assume that it is less than 231.

Example

Input

24 3
24 2
2 1
1 1
4 2
18 3
17 1
17 3
17 4
100 5
1000 10
1120 14
0 0


Output

2
3
1
0
0
2
1
0
1
55
200102899
2079324314

### Response

```cpp
/*
  file : /main.comp.cpp
  date : 2015-10-23

 */

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<cassert>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;
typedef ll int__;
#define rep(i,j) for(int__ i=0;i<(int__)(j);i++)
#define repeat(i,j,k) for(int__ i=(j);i<(int__)(k);i++)
#define all(v) v.begin(),v.end()

#define N_MAX 1130
#define K_MAX 15
int DP[N_MAX][K_MAX];//n k


void init(){
    vector<bool> is_prime(N_MAX,true);
    is_prime[0]=false;
    is_prime[1]=false;
    for(int i=2;i<N_MAX;i++){
        if(not is_prime[i])continue;
        for(int j =i*2;j<N_MAX;j+=i){
            is_prime[j]=false;
        }
    }

    rep(i,N_MAX){
        rep(j,K_MAX){
            DP[i][j]=0;
        }
    }

    // rep(p,N_MAX){
    //     if(is_prime[p])DP[p][1]=1;
    // }
    DP[0][0]=1;
    rep(p,N_MAX){
        if(is_prime[p]){
            for(int k=K_MAX-1;k>=0;k--){
                rep(n,N_MAX){
                    if(n-p>=0 and k-1>=0)DP[n][k] += DP[n-p][k-1];
                    //printf("%lld %d %d\n",n,k,DP[n][k]);
                }
            }
        }


    }
}


bool solve(){
    int n,k;
    cin>>n>>k;
    if(n==0 && k==0)return false;
    cout << DP[n][k] <<endl;    
    return true;
}

int main()
{
    init();

    // rep(n,25){
    //     cout<<n<<"  ";
    //     rep(k,3){
    //         cout <<DP[n][k] <<" ";
    //     }
    //     cout <<endl;
    // }
    
    while(solve());
    return 0;
}
```