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

const int NMAX=1120, JMAX=187, KMAX=14;
int dp[NMAX+1][JMAX+1][KMAX+1];
vector<int> primes;

void init(){
	for(int i=0; i<=NMAX; ++i)for(int j=0; j<=JMAX; ++j)for(int k=0; k<=KMAX; ++k)dp[i][j][k]=0;

	bool isPrime[NMAX+1];
	fill(begin(isPrime), end(isPrime), true);
	isPrime[0]=isPrime[1]=false;
	for(int i=2; i<=NMAX; ++i){
		if(isPrime[i]){
			dp[i][primes.size()][1]=1;
			primes.push_back(i);
			for(int j=i+i; j<=NMAX; j+=i)	isPrime[j]=false;
		}
	}
//	cout << primes.size() << endl;

	for(int k=1; k<KMAX; ++k){
		for(int n=2; n<=NMAX; ++n){
			for(int j=0; j<primes.size(); ++j){
				for(int j2=j+1; j2<primes.size(); ++j2){
					if(n+primes[j2]>NMAX)	break;
					dp[n+primes[j2]][j2][k+1] += dp[n][j][k];
				}
			}
		}
	}

}

int main(){
	init();

	int N,K;
	while(true){
		cin>>N>>K;
		if((N|K)==0)	break;
		int ans=0;
		for(int j=0; j<primes.size(); ++j)	ans+=dp[N][j][K];
		cout<<ans<<endl;
	}

//	for(int j=0; j<pries.size(); ++j){
//		cout << j << " " << dp[2][j][2] << endl;
//	}

//	for(int i=1; i<=NMAX; ++i)for(int k=1; k<=KMAX; ++k){
//		int tmp=0;
//		for(int j=0; j<primes.size(); ++j)	tmp+=dp[i][j][k];
//		cout<<i<<" "<<k<<" "<<tmp<<endl;
//	}

	return 0;
}
### Prompt

Please create a solution in Cpp 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 <vector>
#include <algorithm>
using namespace std;

const int NMAX=1120, JMAX=187, KMAX=14;
int dp[NMAX+1][JMAX+1][KMAX+1];
vector<int> primes;

void init(){
	for(int i=0; i<=NMAX; ++i)for(int j=0; j<=JMAX; ++j)for(int k=0; k<=KMAX; ++k)dp[i][j][k]=0;

	bool isPrime[NMAX+1];
	fill(begin(isPrime), end(isPrime), true);
	isPrime[0]=isPrime[1]=false;
	for(int i=2; i<=NMAX; ++i){
		if(isPrime[i]){
			dp[i][primes.size()][1]=1;
			primes.push_back(i);
			for(int j=i+i; j<=NMAX; j+=i)	isPrime[j]=false;
		}
	}
//	cout << primes.size() << endl;

	for(int k=1; k<KMAX; ++k){
		for(int n=2; n<=NMAX; ++n){
			for(int j=0; j<primes.size(); ++j){
				for(int j2=j+1; j2<primes.size(); ++j2){
					if(n+primes[j2]>NMAX)	break;
					dp[n+primes[j2]][j2][k+1] += dp[n][j][k];
				}
			}
		}
	}

}

int main(){
	init();

	int N,K;
	while(true){
		cin>>N>>K;
		if((N|K)==0)	break;
		int ans=0;
		for(int j=0; j<primes.size(); ++j)	ans+=dp[N][j][K];
		cout<<ans<<endl;
	}

//	for(int j=0; j<pries.size(); ++j){
//		cout << j << " " << dp[2][j][2] << endl;
//	}

//	for(int i=1; i<=NMAX; ++i)for(int k=1; k<=KMAX; ++k){
//		int tmp=0;
//		for(int j=0; j<primes.size(); ++j)	tmp+=dp[i][j][k];
//		cout<<i<<" "<<k<<" "<<tmp<<endl;
//	}

	return 0;
}
```