#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

#define MAX_N 1120
#define MAX_K 15

int main(){
  bool p[MAX_N+1];
  memset(p, true, sizeof(p));
  p[0] = p[1] = false;
  for(int i = 2 ; i*i <= MAX_N ; i++){
    if(p[i]){
      for(int j = i*i ; j <= MAX_N ; j += i){
	p[j] = false;
      }
    }
  }
  vector<int> prime;
  for(int i = 0 ; i <= MAX_N ; i++){
    if(p[i]) prime.push_back(i);
  }
  
  int dp[MAX_K+1][MAX_N+1]; // dp[used number][sum] = number of ways
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  
  for(int i = 0 ; i < prime.size() ; i++){
    for(int j = MAX_N ; j >= prime[i] ; j--){
      for(int k = 14 ; k > 0 ; k--){
	dp[k][j] += dp[k-1][j - prime[i]];
      }
    }
  }

  int n, m;
  while(cin >> n >> m, n|m){
    cout << dp[m][n] << endl;
  }
  
  return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

#define MAX_N 1120
#define MAX_K 15

int main(){
  bool p[MAX_N+1];
  memset(p, true, sizeof(p));
  p[0] = p[1] = false;
  for(int i = 2 ; i*i <= MAX_N ; i++){
    if(p[i]){
      for(int j = i*i ; j <= MAX_N ; j += i){
	p[j] = false;
      }
    }
  }
  vector<int> prime;
  for(int i = 0 ; i <= MAX_N ; i++){
    if(p[i]) prime.push_back(i);
  }
  
  int dp[MAX_K+1][MAX_N+1]; // dp[used number][sum] = number of ways
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  
  for(int i = 0 ; i < prime.size() ; i++){
    for(int j = MAX_N ; j >= prime[i] ; j--){
      for(int k = 14 ; k > 0 ; k--){
	dp[k][j] += dp[k-1][j - prime[i]];
      }
    }
  }

  int n, m;
  while(cin >> n >> m, n|m){
    cout << dp[m][n] << endl;
  }
  
  return 0;
}
```