#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

#include <iostream>
#include <complex>
#include <string>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>

#include <functional>
#include <cassert>

typedef long long ll;
using namespace std;

#ifndef LOCAL
#define debug(x) ;
#else
#define debug(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl;

template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}

template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << '{';
  for (const T &item : v) out << item << ", ";
  out << "\b\b}";
  return out;
}
#endif

#define mod 1000000007 //1e9+7(prime number)
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define SIZE 200010

vector<bool> isPrime;

void Eratosthenes(int N){
  isPrime.assign(N+1, true);
  isPrime[0] = isPrime[1] = false;

  for(int i=2; i*i<=N; i++)
    if(isPrime[i])
      for(int j=i; i*j<=N; j++)
        isPrime[i*j] = false;
}

int dp[15][1121] = {};

bool solve() {
  int N, K;

  scanf("%d%d", &N, &K);

  if (N == 0) return false;

  memset(dp, 0, sizeof(dp));

  dp[0][0] = 1;

  for(int i=0; i<=N; i++) {
    if (!isPrime[i]) continue;

    for(int k=K-1; k>=0; k--) {
      for(int j=0; j+i<=N; j++) {
        dp[k+1][j+i] += dp[k][j];
      }
    }
  }

  printf("%d\n", dp[K][N]);

  return true;
}

int main(){

  Eratosthenes(1120);

  while(solve());

  return 0;
}


### Prompt

Your task is to create 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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

#include <iostream>
#include <complex>
#include <string>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>

#include <functional>
#include <cassert>

typedef long long ll;
using namespace std;

#ifndef LOCAL
#define debug(x) ;
#else
#define debug(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl;

template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}

template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << '{';
  for (const T &item : v) out << item << ", ";
  out << "\b\b}";
  return out;
}
#endif

#define mod 1000000007 //1e9+7(prime number)
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define SIZE 200010

vector<bool> isPrime;

void Eratosthenes(int N){
  isPrime.assign(N+1, true);
  isPrime[0] = isPrime[1] = false;

  for(int i=2; i*i<=N; i++)
    if(isPrime[i])
      for(int j=i; i*j<=N; j++)
        isPrime[i*j] = false;
}

int dp[15][1121] = {};

bool solve() {
  int N, K;

  scanf("%d%d", &N, &K);

  if (N == 0) return false;

  memset(dp, 0, sizeof(dp));

  dp[0][0] = 1;

  for(int i=0; i<=N; i++) {
    if (!isPrime[i]) continue;

    for(int k=K-1; k>=0; k--) {
      for(int j=0; j+i<=N; j++) {
        dp[k+1][j+i] += dp[k][j];
      }
    }
  }

  printf("%d\n", dp[K][N]);

  return true;
}

int main(){

  Eratosthenes(1120);

  while(solve());

  return 0;
}


```