#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>

using namespace std;

const int MAX = 1200;
const int K_MAX = 15;
vector<int> Prime;
int dp[MAX][K_MAX][200];

void make(){
  int isPrime[MAX];
  memset(isPrime,true,sizeof(isPrime));
  for(int i = 2; i*i < MAX; i++)
    if(isPrime[i]) for(int j = 2; i*j < MAX; j++) isPrime[i*j]=false;

  for(int i = 2; i < MAX; i++)
    if(isPrime[i]) Prime.push_back(i);
}

void solve(){
  memset(dp,0,sizeof(dp));
  dp[0][0][0] = 1;
  
  for(int j = 1; j < K_MAX; j++)
    for(int k = 0; k < (int)Prime.size(); k++)
      for(int i = 0; i < MAX; i++)
	for(int x = 0; x <= k; x++){
	  if(j != 1 && x == k) break;
	  if(i-Prime[k] >= 0) dp[i][j][k] += dp[i-Prime[k]][j-1][x];
	}
}

int main(){
  make();
  solve();
  int n,k;
  while(cin >> n >> k && n+k){
    int sum = 0;
    for(int i = 0; i < 200; i++) sum += dp[n][k][i];
    cout << sum << endl;
  }
  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<vector>
#include<algorithm>
#include<cstring>

using namespace std;

const int MAX = 1200;
const int K_MAX = 15;
vector<int> Prime;
int dp[MAX][K_MAX][200];

void make(){
  int isPrime[MAX];
  memset(isPrime,true,sizeof(isPrime));
  for(int i = 2; i*i < MAX; i++)
    if(isPrime[i]) for(int j = 2; i*j < MAX; j++) isPrime[i*j]=false;

  for(int i = 2; i < MAX; i++)
    if(isPrime[i]) Prime.push_back(i);
}

void solve(){
  memset(dp,0,sizeof(dp));
  dp[0][0][0] = 1;
  
  for(int j = 1; j < K_MAX; j++)
    for(int k = 0; k < (int)Prime.size(); k++)
      for(int i = 0; i < MAX; i++)
	for(int x = 0; x <= k; x++){
	  if(j != 1 && x == k) break;
	  if(i-Prime[k] >= 0) dp[i][j][k] += dp[i-Prime[k]][j-1][x];
	}
}

int main(){
  make();
  solve();
  int n,k;
  while(cin >> n >> k && n+k){
    int sum = 0;
    for(int i = 0; i < 200; i++) sum += dp[n][k][i];
    cout << sum << endl;
  }
  return 0;
}
```