#include<iostream>
#include<vector>
#include<algorithm>
#include<string.h>
using namespace std;


vector<int> sieve_of_eratosthenes(int n) {
  vector<int> primes(n);
  for (int i = 2; i < n; ++i)
    primes[i] = i;
  for (int i = 2; i*i < n; ++i)
    if (primes[i])
      for (int j = i*i; j < n; j+=i)
        primes[j] = 0;
  return primes;
}

		int dp[15][200][1121];

int main()
{
	vector<int> a=sieve_of_eratosthenes(1121);
	vector<int> prime;
	for(int i=0;i<a.size();i++)
		if(a[i]!=0) prime.push_back(a[i]);

	int n,k;
	while(cin>>n>>k && n!=0){
		memset(dp,0,sizeof(dp));

		dp[0][0][0]=1;

		for(int i=0;i<k;i++){
			for(int j=0;j<prime.size();j++){
				for(int num=0;num<=n;num++){
					if(dp[i][j][num]==0) continue;
					for(int next=j;next<prime.size();next++){
						if(num+prime[next]>n) break;
						dp[i+1][next+1][num+prime[next]]+=dp[i][j][num];
					}
				}
			}

		}
		int ans=0;
		for(int i=0;i<=prime.size();i++){
			ans+=dp[k][i][n];
		}
		cout<<ans<<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<string.h>
using namespace std;


vector<int> sieve_of_eratosthenes(int n) {
  vector<int> primes(n);
  for (int i = 2; i < n; ++i)
    primes[i] = i;
  for (int i = 2; i*i < n; ++i)
    if (primes[i])
      for (int j = i*i; j < n; j+=i)
        primes[j] = 0;
  return primes;
}

		int dp[15][200][1121];

int main()
{
	vector<int> a=sieve_of_eratosthenes(1121);
	vector<int> prime;
	for(int i=0;i<a.size();i++)
		if(a[i]!=0) prime.push_back(a[i]);

	int n,k;
	while(cin>>n>>k && n!=0){
		memset(dp,0,sizeof(dp));

		dp[0][0][0]=1;

		for(int i=0;i<k;i++){
			for(int j=0;j<prime.size();j++){
				for(int num=0;num<=n;num++){
					if(dp[i][j][num]==0) continue;
					for(int next=j;next<prime.size();next++){
						if(num+prime[next]>n) break;
						dp[i+1][next+1][num+prime[next]]+=dp[i][j][num];
					}
				}
			}

		}
		int ans=0;
		for(int i=0;i<=prime.size();i++){
			ans+=dp[k][i][n];
		}
		cout<<ans<<endl;
	}

	return 0;
}
```