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

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define DEB 0
#define all(x) x.begin(), x.end()

int sz;
int primes[1120];
bool check[5000];
int dp[190][1120][15];

void make_prime(){
  check[0]=check[1]=true;
  for(int i=2; i*i<=1120; i++){
    if( !check[i] ){
      for(int j=i*2; j<=1120; j+=i){
	check[j] = true;
      }
    }
  }
  for(int i=2; i<=1120; i++){
    if( !check[i] ){
      primes[sz++] = i;
    }
  }
}

int solve(int idx, int n, int k){
  int ret = 0;
#if DEB
  printf("(%d,%d,%d)\n",idx,n,k);
#endif
  if( dp[idx][n][k]!=-1 ){
    return dp[idx][n][k];
  }
  if( k==0 ){
    return n==0;
  }
  for(int i=idx; i<sz && primes[i]<=n; i++){
    ret += solve(i+1,n-primes[i],k-1);
  }
  return dp[idx][n][k] = ret;
}

int main(){
  int n,k;
  make_prime();
  memset(dp,-1,sizeof(dp));
  while(scanf("%d%d",&n,&k),n|k){
    printf("%d\n",solve(0,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 <algorithm>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define DEB 0
#define all(x) x.begin(), x.end()

int sz;
int primes[1120];
bool check[5000];
int dp[190][1120][15];

void make_prime(){
  check[0]=check[1]=true;
  for(int i=2; i*i<=1120; i++){
    if( !check[i] ){
      for(int j=i*2; j<=1120; j+=i){
	check[j] = true;
      }
    }
  }
  for(int i=2; i<=1120; i++){
    if( !check[i] ){
      primes[sz++] = i;
    }
  }
}

int solve(int idx, int n, int k){
  int ret = 0;
#if DEB
  printf("(%d,%d,%d)\n",idx,n,k);
#endif
  if( dp[idx][n][k]!=-1 ){
    return dp[idx][n][k];
  }
  if( k==0 ){
    return n==0;
  }
  for(int i=idx; i<sz && primes[i]<=n; i++){
    ret += solve(i+1,n-primes[i],k-1);
  }
  return dp[idx][n][k] = ret;
}

int main(){
  int n,k;
  make_prime();
  memset(dp,-1,sizeof(dp));
  while(scanf("%d%d",&n,&k),n|k){
    printf("%d\n",solve(0,n,k));
  }
  return 0;
}
```