#include <algorithm>
#include <vector>
#include <cfloat>
#include <string>
#include <cmath>
#include <set>
#include <cstdlib>
#include <map>
#include <ctime>
#include <iomanip>
#include <functional>
#include <deque>
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
#include <stack>
#include <climits>
#include <sys/time.h>
#include <cctype>

using namespace std;

typedef long long ll;

ll memo[200][200];
ll combi(ll a, ll b) {
  if (memo[a][b]) return memo[a][b];
  if (b == 1) return a;
  if (a == b) return 1;
  return memo[a][b] = combi(a-1, b-1)+combi(a-1, b);
}

int main() {
  bool notPrime[1121];
  memset(notPrime, false, sizeof(notPrime));
  for (int i = 2; i*i <= 1120; i++) {
    for (int j = i+i; j <= 1120; j+=i) {
      notPrime[j] = true;
    }
  }
  vector <int> prime;
  for (int i = 2; i <= 1120; i++) {
    if (!notPrime[i]) prime.push_back(i);
  }
  int m = prime.size();
  while (true) {
    int n, k;
    cin >> n >> k;
    if (n == 0 && k == 0) break;
    // cout << 1121 << " " << m << " " << k+1 << " " << ((1LL<<31)-1) << endl;
    ll dp[1121][m][2];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < m; i++) {
      dp[prime[i]][i][1] = 1;
    }

    // for (int p = 0; prime[p] < n; p++) {	
    //   cout << prime[p] << " ";
    // }
    // cout << endl;
    for (int i = 2; i <= k; i++) {
      for (int t1 = 0; t1 <= n; t1++) {
	for (int t2 = 0; t2 < m; t2++) {
	  dp[t1][t2][i%2] = 0;
	}
      }
      for (int p1 = 0; p1 < m; p1++) {
	for (int num = 2; num < n; num++) {
	  if (dp[num][p1][(i+1)%2] == 0) continue;
	  for (int p2 = p1+1; p2 < m; p2++) {
	    if (num+prime[p2] > n) break;

	    dp[num+prime[p2]][p2][i%2] += dp[num][p1][(i+1)%2];
	  }
	}
      }
      // for (int p = 0; prime[p] < n; p++) {
      // 	cout << dp[n][p][i%2] << " ";
      // }
      // cout << endl;
      // for (int t1 = 0; t1 <= n; t1++) {
      // 	cout << setw(2) << t1 << ":";
      // 	for (int t2 = 0; t2 < m && prime[t2] <= n; t2++) {
      // 	  cout << dp[t1][t2][i%2] << " ";
      // 	}
      // 	cout << endl;
      // }
      // cout << endl;
    }
    ll ans = 0;
    for (int i = 0; i < m; i++) ans += dp[n][i][k%2];
    cout << ans << endl;
  }
}
### 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 <algorithm>
#include <vector>
#include <cfloat>
#include <string>
#include <cmath>
#include <set>
#include <cstdlib>
#include <map>
#include <ctime>
#include <iomanip>
#include <functional>
#include <deque>
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
#include <stack>
#include <climits>
#include <sys/time.h>
#include <cctype>

using namespace std;

typedef long long ll;

ll memo[200][200];
ll combi(ll a, ll b) {
  if (memo[a][b]) return memo[a][b];
  if (b == 1) return a;
  if (a == b) return 1;
  return memo[a][b] = combi(a-1, b-1)+combi(a-1, b);
}

int main() {
  bool notPrime[1121];
  memset(notPrime, false, sizeof(notPrime));
  for (int i = 2; i*i <= 1120; i++) {
    for (int j = i+i; j <= 1120; j+=i) {
      notPrime[j] = true;
    }
  }
  vector <int> prime;
  for (int i = 2; i <= 1120; i++) {
    if (!notPrime[i]) prime.push_back(i);
  }
  int m = prime.size();
  while (true) {
    int n, k;
    cin >> n >> k;
    if (n == 0 && k == 0) break;
    // cout << 1121 << " " << m << " " << k+1 << " " << ((1LL<<31)-1) << endl;
    ll dp[1121][m][2];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < m; i++) {
      dp[prime[i]][i][1] = 1;
    }

    // for (int p = 0; prime[p] < n; p++) {	
    //   cout << prime[p] << " ";
    // }
    // cout << endl;
    for (int i = 2; i <= k; i++) {
      for (int t1 = 0; t1 <= n; t1++) {
	for (int t2 = 0; t2 < m; t2++) {
	  dp[t1][t2][i%2] = 0;
	}
      }
      for (int p1 = 0; p1 < m; p1++) {
	for (int num = 2; num < n; num++) {
	  if (dp[num][p1][(i+1)%2] == 0) continue;
	  for (int p2 = p1+1; p2 < m; p2++) {
	    if (num+prime[p2] > n) break;

	    dp[num+prime[p2]][p2][i%2] += dp[num][p1][(i+1)%2];
	  }
	}
      }
      // for (int p = 0; prime[p] < n; p++) {
      // 	cout << dp[n][p][i%2] << " ";
      // }
      // cout << endl;
      // for (int t1 = 0; t1 <= n; t1++) {
      // 	cout << setw(2) << t1 << ":";
      // 	for (int t2 = 0; t2 < m && prime[t2] <= n; t2++) {
      // 	  cout << dp[t1][t2][i%2] << " ";
      // 	}
      // 	cout << endl;
      // }
      // cout << endl;
    }
    ll ans = 0;
    for (int i = 0; i < m; i++) ans += dp[n][i][k%2];
    cout << ans << endl;
  }
}
```