#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

bool isPrime(int n) {
  if( n < 2 ) return false;
  if( n == 2 ) return true;
  if( n % 2 == 0 ) return false;
  for(int d = 3; d * d <= n; d += 2 ) {
    if( n % d == 0 ) return false;
  }
  return true;
}

vector<int> primes;

int dp[1500][16];

int main() {
  for(int i = 0; i < 1120; i++)
    if( isPrime(i) )
      primes.push_back(i);

  int N = primes.size();

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

  for(int i = 0; i < N; i++){
    for(int k = 14; k >= 0; k--) {
      for(int n = 1; n <= 1120; n++) {
        if( k == 1 && n - primes[i] == 0 ) {
          dp[n][k]++;
        }

        if( n - primes[i] > 0 ) {
          dp[n][k] += dp[n-primes[i]][k-1];
        }
      }
    }
  }

  for( int n, k; scanf("%d%d", &n, &k), (n || k); ) {
    printf("%d\n", dp[n][k]);
  }
  return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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<cstring>
using namespace std;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

bool isPrime(int n) {
  if( n < 2 ) return false;
  if( n == 2 ) return true;
  if( n % 2 == 0 ) return false;
  for(int d = 3; d * d <= n; d += 2 ) {
    if( n % d == 0 ) return false;
  }
  return true;
}

vector<int> primes;

int dp[1500][16];

int main() {
  for(int i = 0; i < 1120; i++)
    if( isPrime(i) )
      primes.push_back(i);

  int N = primes.size();

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

  for(int i = 0; i < N; i++){
    for(int k = 14; k >= 0; k--) {
      for(int n = 1; n <= 1120; n++) {
        if( k == 1 && n - primes[i] == 0 ) {
          dp[n][k]++;
        }

        if( n - primes[i] > 0 ) {
          dp[n][k] += dp[n-primes[i]][k-1];
        }
      }
    }
  }

  for( int n, k; scanf("%d%d", &n, &k), (n || k); ) {
    printf("%d\n", dp[n][k]);
  }
  return 0;
}
```