#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>
  
using namespace std;
  
typedef long long ll;
typedef pair <int,int> P;
typedef pair <int,P> PP;
  
static const double EPS = 1e-8;
  
static const int tx[] = {0,1,0,-1};
static const int ty[] = {-1,0,1,0};

bool is_prime[1121];
int dp[40001][20]; //dp[num][size] = count

int main(){

  memset(is_prime,true,sizeof(is_prime));

  is_prime[0] = false;
  is_prime[1] = false;
  for(int i=2;i*i<=1120;i++){
    if(!is_prime[i]) continue;

    for(int j=i+i;j<=1120;j+=i){
      is_prime[j] = false;
    }
  }

  vector<int> primes;
  for(int i=0;i<=1120;i++){
    if(is_prime[i]) {
      primes.push_back(i);
    }
  }

  memset(dp,0,sizeof(dp));
  dp[0][0] = 1;
  for(int i=0;i<primes.size();i++){
    for(int num=1120*14;num>=0;num--){
      for(int j=14;j>=0;j--){
	dp[num + primes[i]][j+1] += dp[num][j];
      }
    }
  }
  
  int target;
  int size;
  while(~scanf("%d %d",&target,&size)){
    if(target == 0 && size == 0) break;
    printf("%d\n",dp[target][size]);
  }
}
### Prompt

Create a solution in cpp for 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
#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>
  
using namespace std;
  
typedef long long ll;
typedef pair <int,int> P;
typedef pair <int,P> PP;
  
static const double EPS = 1e-8;
  
static const int tx[] = {0,1,0,-1};
static const int ty[] = {-1,0,1,0};

bool is_prime[1121];
int dp[40001][20]; //dp[num][size] = count

int main(){

  memset(is_prime,true,sizeof(is_prime));

  is_prime[0] = false;
  is_prime[1] = false;
  for(int i=2;i*i<=1120;i++){
    if(!is_prime[i]) continue;

    for(int j=i+i;j<=1120;j+=i){
      is_prime[j] = false;
    }
  }

  vector<int> primes;
  for(int i=0;i<=1120;i++){
    if(is_prime[i]) {
      primes.push_back(i);
    }
  }

  memset(dp,0,sizeof(dp));
  dp[0][0] = 1;
  for(int i=0;i<primes.size();i++){
    for(int num=1120*14;num>=0;num--){
      for(int j=14;j>=0;j--){
	dp[num + primes[i]][j+1] += dp[num][j];
      }
    }
  }
  
  int target;
  int size;
  while(~scanf("%d %d",&target,&size)){
    if(target == 0 && size == 0) break;
    printf("%d\n",dp[target][size]);
  }
}
```