#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
#define pb push_back
int n, k;
const int MAX_N = 1120, MAX_K = 14;
vector<int> primes;
int dp[MAX_N+1][MAX_K+1][200];

int rec(int an, int ak, int pidx) {
  if (an == 0 && ak == 0) return 1;
  if (an == 0 || ak == 0) return 0;
  
  if (dp[an][ak][pidx] >= 0) return dp[an][ak][pidx];
  
  // ツ素ツ青板づーツ使ツつ、
  int cnt = 0;
  for (int i = pidx; i < primes.size(); i += 1) {
    if (primes[i] > an) break;
    
    cnt += rec(an - primes[i], ak - 1, i+1);
  }
  return dp[an][ak][pidx] = cnt;
}

void solve() {
  memset(dp, -1, sizeof dp);
  cout<<rec(n, k, 0)<<endl;
}

void set_prime_list() {
  bool sieve[MAX_N];
  memset(sieve, true, sizeof sieve);
  
  for(int i=2;i<MAX_N;i++)
  if (sieve[i]) {
    primes.pb(i);
    for(int j=i+i;j<MAX_N;j+=i)
    sieve[j] = false;
  }
}

int main (int argc, char const* argv[]) {
  set_prime_list();
  
  while (cin>>n>>k, n||k) {
    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 <iostream>
#include <vector>
#include <cstring>
using namespace std;
#define pb push_back
int n, k;
const int MAX_N = 1120, MAX_K = 14;
vector<int> primes;
int dp[MAX_N+1][MAX_K+1][200];

int rec(int an, int ak, int pidx) {
  if (an == 0 && ak == 0) return 1;
  if (an == 0 || ak == 0) return 0;
  
  if (dp[an][ak][pidx] >= 0) return dp[an][ak][pidx];
  
  // ツ素ツ青板づーツ使ツつ、
  int cnt = 0;
  for (int i = pidx; i < primes.size(); i += 1) {
    if (primes[i] > an) break;
    
    cnt += rec(an - primes[i], ak - 1, i+1);
  }
  return dp[an][ak][pidx] = cnt;
}

void solve() {
  memset(dp, -1, sizeof dp);
  cout<<rec(n, k, 0)<<endl;
}

void set_prime_list() {
  bool sieve[MAX_N];
  memset(sieve, true, sizeof sieve);
  
  for(int i=2;i<MAX_N;i++)
  if (sieve[i]) {
    primes.pb(i);
    for(int j=i+i;j<MAX_N;j+=i)
    sieve[j] = false;
  }
}

int main (int argc, char const* argv[]) {
  set_prime_list();
  
  while (cin>>n>>k, n||k) {
    solve();
  }
  return 0;
}
```