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

using namespace std;

typedef long long int lli;

lli memo[1121][15][188];
vector<int> prm;
int l;

lli func(int n, int k, int p, int t, int c)
{
  if(n == t && k == c){
    return 1;
  }
  if(n == t){
    return 0;
  }
  if(k == c){
    return 0;
  }
  if(p == l){
    return 0;
  }
  if(memo[t][c][p] >= 0){
    return memo[t][c][p];
  }
  lli u = 0;
  for(int i = p; i < l; i++){
    if((t+prm[i]) > n){
      break;
    }
    u += func(n, k, i+1, t+prm[i], c+1);
  }
  memo[t][c][p] = u;

  return u;
}

int main()
{
  int n, k;
  bool a[1121];
  fill(a, a+1121, true);
  a[0] = a[1] = false;

  for(int i = 2; i < 1121; i++){
    if(a[i]){
      prm.push_back(i);
      for(int j = i+i; j < 1121; j+=i){
	a[j] = false;
      }
    }
  }
  l = prm.size();

  while(cin>>n>>k && (k+n)){
    fill(&memo[0][0][0], &memo[1120][14][187], -1);
    cout << func(n, k, 0, 0, 0) << endl;
  }

  return 0;
}
### Prompt

Develop a solution in Cpp 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 <algorithm>
#include <vector>

using namespace std;

typedef long long int lli;

lli memo[1121][15][188];
vector<int> prm;
int l;

lli func(int n, int k, int p, int t, int c)
{
  if(n == t && k == c){
    return 1;
  }
  if(n == t){
    return 0;
  }
  if(k == c){
    return 0;
  }
  if(p == l){
    return 0;
  }
  if(memo[t][c][p] >= 0){
    return memo[t][c][p];
  }
  lli u = 0;
  for(int i = p; i < l; i++){
    if((t+prm[i]) > n){
      break;
    }
    u += func(n, k, i+1, t+prm[i], c+1);
  }
  memo[t][c][p] = u;

  return u;
}

int main()
{
  int n, k;
  bool a[1121];
  fill(a, a+1121, true);
  a[0] = a[1] = false;

  for(int i = 2; i < 1121; i++){
    if(a[i]){
      prm.push_back(i);
      for(int j = i+i; j < 1121; j+=i){
	a[j] = false;
      }
    }
  }
  l = prm.size();

  while(cin>>n>>k && (k+n)){
    fill(&memo[0][0][0], &memo[1120][14][187], -1);
    cout << func(n, k, 0, 0, 0) << endl;
  }

  return 0;
}
```