#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

const int K = 14 + 1;
const int P = 190;
const int N = 1120 + 1;

int t[K][P][N];

int p[P], size = 0;
bool prime[N];

int result[K][N];

int main(void)
{  
  fill( prime, prime + N, true );
  prime[0] = prime[1] = false;
  for(int i=2; i*i<N; ++i){
    for(int j=2; i*j<N; ++j){
      prime[i*j] = false;
    }
  }
  for(int i=0; i<N; ++i){
    if( prime[i] )p[size++] = i;
  }

  for(int i=0; i<size; ++i){
    t[1][i][p[i]] = 1;
  }

  for(int i=0; i+1<K; ++i){
    for(int j=0; j<size; ++j){
      for(int k=0; k<N; ++k){
	if( t[i][j][k] == 0 )continue;
	for(int l=j+1; l<size; ++l){
	  if(k + p[l] < N){
	    t[i+1][l][k+p[l]] += t[i][j][k];
	  }
	}
      }
    }
  }

  for(int i=0; i<K; ++i){
    for(int j=0; j<P; ++j){
      for(int k=0; k<N; ++k){
	result[i][k] += t[i][j][k];
      }
    }
  }

  for(int n, k; scanf("%d %d\n", &n, &k) && n && k; printf("%d\n", result[k][n]));

  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 <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

const int K = 14 + 1;
const int P = 190;
const int N = 1120 + 1;

int t[K][P][N];

int p[P], size = 0;
bool prime[N];

int result[K][N];

int main(void)
{  
  fill( prime, prime + N, true );
  prime[0] = prime[1] = false;
  for(int i=2; i*i<N; ++i){
    for(int j=2; i*j<N; ++j){
      prime[i*j] = false;
    }
  }
  for(int i=0; i<N; ++i){
    if( prime[i] )p[size++] = i;
  }

  for(int i=0; i<size; ++i){
    t[1][i][p[i]] = 1;
  }

  for(int i=0; i+1<K; ++i){
    for(int j=0; j<size; ++j){
      for(int k=0; k<N; ++k){
	if( t[i][j][k] == 0 )continue;
	for(int l=j+1; l<size; ++l){
	  if(k + p[l] < N){
	    t[i+1][l][k+p[l]] += t[i][j][k];
	  }
	}
      }
    }
  }

  for(int i=0; i<K; ++i){
    for(int j=0; j<P; ++j){
      for(int k=0; k<N; ++k){
	result[i][k] += t[i][j][k];
      }
    }
  }

  for(int n, k; scanf("%d %d\n", &n, &k) && n && k; printf("%d\n", result[k][n]));

  return 0;
}
```