#include <iostream>
#include <sstream>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>

using namespace std;

#define FOR(i,k,n) for(int i=(k); i<(int)n; ++i)
#define REP(i,n) FOR(i,0,n)
#define FORIT(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

template<class T> void debug(T begin, T end){ for(T i = begin; i != end; ++i) cout<<*i<<" "; cout<<endl; }

typedef long long ll;
const int INF = 100000000;
const double EPS = 1e-8;
const int MOD = 1000000007;
typedef pair<int,int> P;
typedef pair<P, int> PP;
vector<int> prime;
const int MAXN = 1121;
int memo[MAXN][15][200];
int dfs(int n, int k, int p){
  if(~memo[n][k][p]) return memo[n][k][p];
  if(k == 0 && n == 0) return 1;
  if(k == 0) return 0;
  int res = 0;
  FOR(i, p, prime.size()){
    if(n - prime[i] < 0) break;
    res += dfs(n - prime[i], k - 1, i + 1);
  }
  return memo[n][k][p] = res;
}

int main(){
  memset(memo,-1,sizeof(memo));
  vector<bool> bprime(MAXN, 1);
  FOR(p, 2, MAXN)if(bprime[p]){
    prime.push_back(p);
    for(int q = p * p; q < MAXN; q += p){
      bprime[q] = false;
    }
  }
  int n,k;
  while(cin>>n>>k,n){
    cout<<dfs(n, k, 0)<<endl;
  }
  return 0;
}
### 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 <sstream>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>

using namespace std;

#define FOR(i,k,n) for(int i=(k); i<(int)n; ++i)
#define REP(i,n) FOR(i,0,n)
#define FORIT(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

template<class T> void debug(T begin, T end){ for(T i = begin; i != end; ++i) cout<<*i<<" "; cout<<endl; }

typedef long long ll;
const int INF = 100000000;
const double EPS = 1e-8;
const int MOD = 1000000007;
typedef pair<int,int> P;
typedef pair<P, int> PP;
vector<int> prime;
const int MAXN = 1121;
int memo[MAXN][15][200];
int dfs(int n, int k, int p){
  if(~memo[n][k][p]) return memo[n][k][p];
  if(k == 0 && n == 0) return 1;
  if(k == 0) return 0;
  int res = 0;
  FOR(i, p, prime.size()){
    if(n - prime[i] < 0) break;
    res += dfs(n - prime[i], k - 1, i + 1);
  }
  return memo[n][k][p] = res;
}

int main(){
  memset(memo,-1,sizeof(memo));
  vector<bool> bprime(MAXN, 1);
  FOR(p, 2, MAXN)if(bprime[p]){
    prime.push_back(p);
    for(int q = p * p; q < MAXN; q += p){
      bprime[q] = false;
    }
  }
  int n,k;
  while(cin>>n>>k,n){
    cout<<dfs(n, k, 0)<<endl;
  }
  return 0;
}
```