#include<iostream>
#include<vector>
#include<algorithm>

#define all(c) (c).begin(),(c).end()

using namespace std;

vector<int>w;

bool fg[2000000];

void sieve(void){
  w.clear();
  w.push_back(-1);

  fill(fg,fg+2000000,false);
  for(int i=2;i*i<2000000;i++){
    if(!fg[i]){
      w.push_back(i);
      for(int j=i*2;j<2000000;j+=i){
        fg[j]=true;
      }
    }
  }
}

int dp[20][300][1130];

int main(void){

  int K,n;

  sieve();

  while(cin >> n >> K,n|K){
    fill(*dp[0],*dp[20],0);
    dp[0][0][0]=1;

    int ans=0;
    int sz=(int)(upper_bound(all(w),n)-w.begin());
    for(int i=0;i<K;i++){
      for(int j=0;j<sz;j++){
        for(int k=0;k<=n;k++){
          for(int l=j+1;l<sz;l++){
            if(k+w[l]<=n){
              dp[i+1][l][k+w[l]]+=dp[i][j][k];
            }
          }
        }
      }
    }

    for(int i=0;i<sz;i++)ans+=dp[K][i][n];

    cout << ans << endl;
  }

  return 0;
}
### Prompt

Generate 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
#include<iostream>
#include<vector>
#include<algorithm>

#define all(c) (c).begin(),(c).end()

using namespace std;

vector<int>w;

bool fg[2000000];

void sieve(void){
  w.clear();
  w.push_back(-1);

  fill(fg,fg+2000000,false);
  for(int i=2;i*i<2000000;i++){
    if(!fg[i]){
      w.push_back(i);
      for(int j=i*2;j<2000000;j+=i){
        fg[j]=true;
      }
    }
  }
}

int dp[20][300][1130];

int main(void){

  int K,n;

  sieve();

  while(cin >> n >> K,n|K){
    fill(*dp[0],*dp[20],0);
    dp[0][0][0]=1;

    int ans=0;
    int sz=(int)(upper_bound(all(w),n)-w.begin());
    for(int i=0;i<K;i++){
      for(int j=0;j<sz;j++){
        for(int k=0;k<=n;k++){
          for(int l=j+1;l<sz;l++){
            if(k+w[l]<=n){
              dp[i+1][l][k+w[l]]+=dp[i][j][k];
            }
          }
        }
      }
    }

    for(int i=0;i<sz;i++)ans+=dp[K][i][n];

    cout << ans << endl;
  }

  return 0;
}
```