#include<iostream>
#include<vector>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
bool p[1121];vector<int> vp;int ca[1121][200][15],s;
int dfs(int n,int mx,int k){
	if(ca[n][mx][k]>=0)return ca[n][mx][k];
	int r=0;
	rep(i,mx){
		if(vp[i]>n)break;
		if(k==1&&vp[i]==n)r+=1;
		else if(k>1)r+=dfs(n-vp[i],i,k-1);
	}
	return ca[n][mx][k]=r;
}
int main(){
	int i,j;
	for(i=2;i*i<1121;i++)if(!p[i])for(j=i+i;j<1121;j+=i)p[j]=1;
	for(i=2;i<1121;i++)if(!p[i])vp.push_back(i);s=vp.size();
	rep(i,1121)rep(j,200)rep(k,15)ca[i][j][k]=-1;
	int n,k; while(cin>>n>>k,n)cout<<dfs(n,s,k)<<endl;
	return 0;
}
### 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
#include<iostream>
#include<vector>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
bool p[1121];vector<int> vp;int ca[1121][200][15],s;
int dfs(int n,int mx,int k){
	if(ca[n][mx][k]>=0)return ca[n][mx][k];
	int r=0;
	rep(i,mx){
		if(vp[i]>n)break;
		if(k==1&&vp[i]==n)r+=1;
		else if(k>1)r+=dfs(n-vp[i],i,k-1);
	}
	return ca[n][mx][k]=r;
}
int main(){
	int i,j;
	for(i=2;i*i<1121;i++)if(!p[i])for(j=i+i;j<1121;j+=i)p[j]=1;
	for(i=2;i<1121;i++)if(!p[i])vp.push_back(i);s=vp.size();
	rep(i,1121)rep(j,200)rep(k,15)ca[i][j][k]=-1;
	int n,k; while(cin>>n>>k,n)cout<<dfs(n,s,k)<<endl;
	return 0;
}
```