#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
#include <cassert>
#include <queue>
#include <set>
#include <map>
#include <valarray>
#include <bitset>
#include <stack>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
const int INF = 1<<29;

int dp[15][1121];
int main() {
  int N = 1300;
  vector<int> prime(N, 1);
  REP(i,N) prime[i] = i;
  prime[0] = prime[1] = 0;
  for (int i=2; i<N; ++i)
    if (prime[i])
      for (int j=i*2; j<N; j+=i)
        prime[j] = 0;
  prime.erase(remove(ALL(prime), 0), prime.end());

  dp[0][0] = 1;
  REP(k, prime.size()) {
    for (int i=13; i>=0; --i) {
      for (int j=1120-prime[k]; j>=0; --j) {
        dp[i+1][j+prime[k]] += dp[i][j];
      }
    }
  }
  int n, k;
  while(cin>>n>>k,n||k) {
    cout << dp[k][n] << endl;
  }
  
}
### Prompt

Generate 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 <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
#include <cassert>
#include <queue>
#include <set>
#include <map>
#include <valarray>
#include <bitset>
#include <stack>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
const int INF = 1<<29;

int dp[15][1121];
int main() {
  int N = 1300;
  vector<int> prime(N, 1);
  REP(i,N) prime[i] = i;
  prime[0] = prime[1] = 0;
  for (int i=2; i<N; ++i)
    if (prime[i])
      for (int j=i*2; j<N; j+=i)
        prime[j] = 0;
  prime.erase(remove(ALL(prime), 0), prime.end());

  dp[0][0] = 1;
  REP(k, prime.size()) {
    for (int i=13; i>=0; --i) {
      for (int j=1120-prime[k]; j>=0; --j) {
        dp[i+1][j+prime[k]] += dp[i][j];
      }
    }
  }
  int n, k;
  while(cin>>n>>k,n||k) {
    cout << dp[k][n] << endl;
  }
  
}
```