#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int MAX_N = 1121;
const int MAX_K = 15;
const int MAX_P = 188;

bool isP[MAX_N];
vector<int> P; // size:187

int T[MAX_K][MAX_P][MAX_N];

void make() {
  fill(isP, isP+MAX_N, true);
  isP[0] = isP[1] = false;
  for(int i = 2; i < MAX_N; ++i) {
    if(isP[i]) {
      P.push_back(i);
      for(int j = i*2; j < MAX_N; j += i) isP[j] = false;
    }
  }
}

int rec(int k, int p, int n) {
  int &res = T[k][p][n];
  if(res == -1) {
    res = 0;
    for(int i = p; i < P.size() && n-P[i] >= 0; ++i) {
      res += rec(k-1, i+1, n-P[i]);
    }
  }
  return res;
}

int main() {
  make();
  for(int i = 0; i < MAX_K; ++i) {
    for(int j = 0; j < MAX_P; ++j) {
      for(int k = 0; k < MAX_N; ++k) {
	if(i == 0 && k == 0) T[i][j][k] = 1;
	else if(i == 0 || k == 0) T[i][j][k] = 0;
	else T[i][j][k] = -1;
      }
    }
  }
  int n, k;
  while((cin >> n >> k) && (n || k)) {
    cout << rec(k, 0, n) << endl;
  }
  return 0;
}
### Prompt

Please formulate 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 <algorithm>
#include <vector>
using namespace std;

const int MAX_N = 1121;
const int MAX_K = 15;
const int MAX_P = 188;

bool isP[MAX_N];
vector<int> P; // size:187

int T[MAX_K][MAX_P][MAX_N];

void make() {
  fill(isP, isP+MAX_N, true);
  isP[0] = isP[1] = false;
  for(int i = 2; i < MAX_N; ++i) {
    if(isP[i]) {
      P.push_back(i);
      for(int j = i*2; j < MAX_N; j += i) isP[j] = false;
    }
  }
}

int rec(int k, int p, int n) {
  int &res = T[k][p][n];
  if(res == -1) {
    res = 0;
    for(int i = p; i < P.size() && n-P[i] >= 0; ++i) {
      res += rec(k-1, i+1, n-P[i]);
    }
  }
  return res;
}

int main() {
  make();
  for(int i = 0; i < MAX_K; ++i) {
    for(int j = 0; j < MAX_P; ++j) {
      for(int k = 0; k < MAX_N; ++k) {
	if(i == 0 && k == 0) T[i][j][k] = 1;
	else if(i == 0 || k == 0) T[i][j][k] = 0;
	else T[i][j][k] = -1;
      }
    }
  }
  int n, k;
  while((cin >> n >> k) && (n || k)) {
    cout << rec(k, 0, n) << endl;
  }
  return 0;
}
```