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

using namespace std;

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

int dp[2][N+1][K+1];
bool is_prime[N+1];
vector<int> prime;

int init(){
  fill(dp[0][0], dp[2][0], 0);
  int t = 0;
  dp[t][0][0] = 1;
  for(int k=0;k<prime.size();k++){
    for(int i=0;i<=N;i++) for(int j=0;j<=K;j++) dp[1-t][i][j] = dp[t][i][j];
    for(int i=0;i<=N;i++){
      if(i+prime[k] > N) continue;
      for(int j=0;j<=K-1;j++){
        if(dp[t][i][j] == 0) continue;
        dp[1-t][i+prime[k]][j+1] += dp[t][i][j];
      }
    }
    t = 1-t;
  }
  return t;
}

void sieve(){
  fill(is_prime, is_prime+N+1, true);
  is_prime[0] = is_prime[1] = false;
  for(int i=2;i<=N;i++){
    if(is_prime[i]){
      prime.push_back(i);
      for(int j=2*i;j<=N;j+=i) is_prime[j] = false;
    }
  }
}

int main(){
  sieve();
  int t = init();
  int n, k;
  while(cin >> n >> k && (n|k)){
    cout << dp[t][n][k] << endl;
  }
  return 0;
}
### Prompt

Create a solution in cpp for 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 dp[2][N+1][K+1];
bool is_prime[N+1];
vector<int> prime;

int init(){
  fill(dp[0][0], dp[2][0], 0);
  int t = 0;
  dp[t][0][0] = 1;
  for(int k=0;k<prime.size();k++){
    for(int i=0;i<=N;i++) for(int j=0;j<=K;j++) dp[1-t][i][j] = dp[t][i][j];
    for(int i=0;i<=N;i++){
      if(i+prime[k] > N) continue;
      for(int j=0;j<=K-1;j++){
        if(dp[t][i][j] == 0) continue;
        dp[1-t][i+prime[k]][j+1] += dp[t][i][j];
      }
    }
    t = 1-t;
  }
  return t;
}

void sieve(){
  fill(is_prime, is_prime+N+1, true);
  is_prime[0] = is_prime[1] = false;
  for(int i=2;i<=N;i++){
    if(is_prime[i]){
      prime.push_back(i);
      for(int j=2*i;j<=N;j+=i) is_prime[j] = false;
    }
  }
}

int main(){
  sieve();
  int t = init();
  int n, k;
  while(cin >> n >> k && (n|k)){
    cout << dp[t][n][k] << endl;
  }
  return 0;
}
```