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

using namespace std;

const int N = 1120;
const int K = 14;

int prime[N];
bool is_prime[N+1];

int sieve(){
  int p =0;
  for(int i=0;i<=N;i++) is_prime[i] = true;
  is_prime[0] = is_prime[1] = false;
  for(int i=2;i<=N;i++){
    if(is_prime[i]){
      prime[p++] = i;
      for(int j=2*i;j<=N;j+=i) is_prime[j] = false;
    }
  }
  return p;
}
//187

int main(){
  int size = sieve();
  int dp[K+1][N+1];
  fill(dp[0], dp[K+1], -1);
  dp[0][0] = 1;
  for(int k=0;k<size;k++){
    for(int i=K-1;i>=0;i--){
      for(int j=N;j>=0;j--){
	if(dp[i][j]!=-1){
	  if(j+prime[k]>N) continue;
	  int &tmp = dp[i+1][j+prime[k]];
	  if(tmp == -1){
	    tmp = dp[i][j];
	  }else{
	    tmp += dp[i][j];
	  }
	}
      }
    }
  }
  int n, k;
  while(cin >> n >> k){
    if(n==0 && k==0) break;
    if(dp[k][n] == -1) cout << 0 << endl;
    else cout << dp[k][n] << endl;
  }
  return 0;
}
### Prompt

Your task is to create 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 <algorithm>
#include <vector>

using namespace std;

const int N = 1120;
const int K = 14;

int prime[N];
bool is_prime[N+1];

int sieve(){
  int p =0;
  for(int i=0;i<=N;i++) is_prime[i] = true;
  is_prime[0] = is_prime[1] = false;
  for(int i=2;i<=N;i++){
    if(is_prime[i]){
      prime[p++] = i;
      for(int j=2*i;j<=N;j+=i) is_prime[j] = false;
    }
  }
  return p;
}
//187

int main(){
  int size = sieve();
  int dp[K+1][N+1];
  fill(dp[0], dp[K+1], -1);
  dp[0][0] = 1;
  for(int k=0;k<size;k++){
    for(int i=K-1;i>=0;i--){
      for(int j=N;j>=0;j--){
	if(dp[i][j]!=-1){
	  if(j+prime[k]>N) continue;
	  int &tmp = dp[i+1][j+prime[k]];
	  if(tmp == -1){
	    tmp = dp[i][j];
	  }else{
	    tmp += dp[i][j];
	  }
	}
      }
    }
  }
  int n, k;
  while(cin >> n >> k){
    if(n==0 && k==0) break;
    if(dp[k][n] == -1) cout << 0 << endl;
    else cout << dp[k][n] << endl;
  }
  return 0;
}
```