/*
問題D：異なる素数の和

・案
１、素数リストを作成し
２、動的計画法
n_max = 1120, k　<=　14
素数リストをはだいたい p=188
全探索するなら　p C k = 188 C 14 > 10^20

k = 1   | k = 2
--------|-------
n=4 p=2 | n = 6
n=3 p=3 | n = 6
n=1 p=5 | n = 6

異なる素数を扱うので扱った素数を記憶しておく
*/

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

int p[1130]; // primes
int prime_num;
int c[1121][15][188]; // メモ化

void calc_primes() {
  p[0] = 2;
  prime_num = 1;
  for (int i=3; i<=1120; i+=2) {
    bool add = true;
    for (int j=0; j<prime_num; j++) {
      if (p[j]*p[j] > i) { break; }
      if (i % p[j] == 0) {
        add = false;
        break;
      }
    }
    if (add) { p[prime_num++] = i; }
  }
}

int solve(int n, int k, int idx) {
  if (n == 0 && k == 0) {
    return 1;
  } else if (n == 0 || k == 0) {
    return 0;
  }
  if (c[n][k][idx] >= 0) { return c[n][k][idx]; }

  int cnt = 0;
  for (int i=idx; i<prime_num; i++) {
    if (p[i] > n) {
      break;
    } else {
      cnt += solve(n - p[i], k - 1, i+1);
    }
  }
  c[n][k][idx] = cnt;
  
  return cnt;
}

int main() {
  calc_primes();
//cout << pnum << endl;
//for (int i=0; i<pnum; i++) { cout << p[i] << ", "; }
  memset(c, -1, sizeof(c));
  
  int n, k;
  while (cin >> n >> k, (n || k)) {
    cout << solve(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
/*
問題D：異なる素数の和

・案
１、素数リストを作成し
２、動的計画法
n_max = 1120, k　<=　14
素数リストをはだいたい p=188
全探索するなら　p C k = 188 C 14 > 10^20

k = 1   | k = 2
--------|-------
n=4 p=2 | n = 6
n=3 p=3 | n = 6
n=1 p=5 | n = 6

異なる素数を扱うので扱った素数を記憶しておく
*/

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

int p[1130]; // primes
int prime_num;
int c[1121][15][188]; // メモ化

void calc_primes() {
  p[0] = 2;
  prime_num = 1;
  for (int i=3; i<=1120; i+=2) {
    bool add = true;
    for (int j=0; j<prime_num; j++) {
      if (p[j]*p[j] > i) { break; }
      if (i % p[j] == 0) {
        add = false;
        break;
      }
    }
    if (add) { p[prime_num++] = i; }
  }
}

int solve(int n, int k, int idx) {
  if (n == 0 && k == 0) {
    return 1;
  } else if (n == 0 || k == 0) {
    return 0;
  }
  if (c[n][k][idx] >= 0) { return c[n][k][idx]; }

  int cnt = 0;
  for (int i=idx; i<prime_num; i++) {
    if (p[i] > n) {
      break;
    } else {
      cnt += solve(n - p[i], k - 1, i+1);
    }
  }
  c[n][k][idx] = cnt;
  
  return cnt;
}

int main() {
  calc_primes();
//cout << pnum << endl;
//for (int i=0; i<pnum; i++) { cout << p[i] << ", "; }
  memset(c, -1, sizeof(c));
  
  int n, k;
  while (cin >> n >> k, (n || k)) {
    cout << solve(n, k, 0) << endl;
  }
  
  return 0;
}

```