#include <iostream>

#define MAXN 1120
#define MAXP 187
#define MAXK 14

using namespace std;

int dp[MAXN+1][MAXP+1][MAXK+1];

int Prime(int p){
  int i,f;
  f = 0;
  if(p == 2) return 1;
  if(p % 2 == 0) return 0;
  for(i=3;i*i <= p;i += 2){
    if(p % i == 0) f = 1;
  }

  if(f == 0) return 1;
  else return 0;
}

int main(){
  int prime[MAXP+1],cnt,n,kk;
  cnt = 0;
  for(int i=2;i <= MAXN;i++){
    if(Prime(i)) prime[cnt++] = i;
  }
  
  for(int i=0;i <= MAXN;i++){
    for(int j=0;j < MAXP;j++){
      for(int k=0;k <= MAXK;k++){
        dp[i][j][k] = 0;
      if(prime[j] == i) dp[i][j][1] = 1;
      }
    }
  }
 for(int j=1;j < MAXP;j++){
  for(int i=0;i<=MAXN;i++){
      for(int k=1;k<=MAXK;k++){
        dp[i][j][k] += dp[i][j-1][k];
        if(i >= prime[j]) dp[i][j][k] += dp[i-prime[j]][j-1][k-1];
      }
    }
  }

  
  while(cin >> n >> kk){
    if(!(n|kk)) break;
    cout << dp[n][MAXP-1][kk] << 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>

#define MAXN 1120
#define MAXP 187
#define MAXK 14

using namespace std;

int dp[MAXN+1][MAXP+1][MAXK+1];

int Prime(int p){
  int i,f;
  f = 0;
  if(p == 2) return 1;
  if(p % 2 == 0) return 0;
  for(i=3;i*i <= p;i += 2){
    if(p % i == 0) f = 1;
  }

  if(f == 0) return 1;
  else return 0;
}

int main(){
  int prime[MAXP+1],cnt,n,kk;
  cnt = 0;
  for(int i=2;i <= MAXN;i++){
    if(Prime(i)) prime[cnt++] = i;
  }
  
  for(int i=0;i <= MAXN;i++){
    for(int j=0;j < MAXP;j++){
      for(int k=0;k <= MAXK;k++){
        dp[i][j][k] = 0;
      if(prime[j] == i) dp[i][j][1] = 1;
      }
    }
  }
 for(int j=1;j < MAXP;j++){
  for(int i=0;i<=MAXN;i++){
      for(int k=1;k<=MAXK;k++){
        dp[i][j][k] += dp[i][j-1][k];
        if(i >= prime[j]) dp[i][j][k] += dp[i-prime[j]][j-1][k-1];
      }
    }
  }

  
  while(cin >> n >> kk){
    if(!(n|kk)) break;
    cout << dp[n][MAXP-1][kk] << endl;
  }

  return 0;
}
```