#include <iostream>
#include <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef complex<double> P;
const int MAX_N = 1130;
const int MAX_K = 17;

bool p[MAX_N];
vector<int> vp;

void make_p(){
  REP(i, MAX_N) p[i] = true;
  p[0] = p[1] = false;
  for(int i = 2; i < MAX_N; ++i){
    if(!p[i]) continue;
    vp.push_back(i);
    for(int j = i + i; j < MAX_N; j += i) p[j] = false;
  }
}

int main() {
  make_p();
  int N, K, dp[MAX_N][MAX_K];
  memset(dp, 0, sizeof(dp));
  REP(i, vp.size()){
    int p = vp[i];
    for(int k = MAX_K - 2; k >= 0; --k){
      REP(n, MAX_N){
        if(p + n >= MAX_N) break;
        dp[n + p][k + 1] += dp[n][k];
      }
    }
    dp[p][1] += 1;
  }
  while(cin >>N >>K && K) cout <<dp[N][K] <<endl;
  return 0;
}
### Prompt

Your challenge is to write 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 <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef complex<double> P;
const int MAX_N = 1130;
const int MAX_K = 17;

bool p[MAX_N];
vector<int> vp;

void make_p(){
  REP(i, MAX_N) p[i] = true;
  p[0] = p[1] = false;
  for(int i = 2; i < MAX_N; ++i){
    if(!p[i]) continue;
    vp.push_back(i);
    for(int j = i + i; j < MAX_N; j += i) p[j] = false;
  }
}

int main() {
  make_p();
  int N, K, dp[MAX_N][MAX_K];
  memset(dp, 0, sizeof(dp));
  REP(i, vp.size()){
    int p = vp[i];
    for(int k = MAX_K - 2; k >= 0; --k){
      REP(n, MAX_N){
        if(p + n >= MAX_N) break;
        dp[n + p][k + 1] += dp[n][k];
      }
    }
    dp[p][1] += 1;
  }
  while(cin >>N >>K && K) cout <<dp[N][K] <<endl;
  return 0;
}
```