#include <bits/stdc++.h>
#define M 2000
using namespace std;
bool used2[M];
vector <int> prime;
int n,k;

int mem[200][15][1200],used[200][15][1200];
int dfs(int idx,int dep,int sum){
  if(sum==n&&dep==k) return 1;
  if(idx==(int)prime.size()||dep==k||sum>=n) return 0;
  if(used[idx][dep][sum]++) return mem[idx][dep][sum];
  
  int res=dfs(idx+1,dep,sum)+dfs(idx+1,dep+1,sum+prime[idx]);
  return mem[idx][dep][sum]=res;
}


int main(){
  used2[0]=used2[1]=1;
  for(int i=2;i*i<M;i++)
    if(!used2[i])
      for(int j=2;j<M/i;j++) used2[j*i]=1;
  
  for(int i=2;i<=1120;i++) if(!used2[i]) prime.push_back(i);
  while(1){
    cin>>n>>k;
    if(n==0&&k==0)break;
    memset(used,0,sizeof(used));
    cout<< dfs(0,0,0)<<endl;
  }
  
  return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
#define M 2000
using namespace std;
bool used2[M];
vector <int> prime;
int n,k;

int mem[200][15][1200],used[200][15][1200];
int dfs(int idx,int dep,int sum){
  if(sum==n&&dep==k) return 1;
  if(idx==(int)prime.size()||dep==k||sum>=n) return 0;
  if(used[idx][dep][sum]++) return mem[idx][dep][sum];
  
  int res=dfs(idx+1,dep,sum)+dfs(idx+1,dep+1,sum+prime[idx]);
  return mem[idx][dep][sum]=res;
}


int main(){
  used2[0]=used2[1]=1;
  for(int i=2;i*i<M;i++)
    if(!used2[i])
      for(int j=2;j<M/i;j++) used2[j*i]=1;
  
  for(int i=2;i<=1120;i++) if(!used2[i]) prime.push_back(i);
  while(1){
    cin>>n>>k;
    if(n==0&&k==0)break;
    memset(used,0,sizeof(used));
    cout<< dfs(0,0,0)<<endl;
  }
  
  return 0;
}
```