#include<iostream>
#include<cstring>
#define pb push_back
#include<vector>
using namespace std;
int n,k;
long long int dp[20][2000];
vector<int> p;
int main(){
	vector<bool> used(1130,true);
	for(int i=2;i<=1120;i++){
		if(used[i]){
			for(int j=i*2;j<=1120;j+=i){
				used[j]=false;
			}
			p.pb(i);
		}
	}
	while(cin>>n>>k,n||k){
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(int i=0;i<p.size()&&p[i]<=n;i++){//i番目の素数
			for(int j=k-1;j>=0;j--){//j回使った
				for(int l=0;l<=n-p[i];l++){//l
					if(dp[j][l]){
						dp[j+1][l+p[i]]+=dp[j][l];
					}
				}
			}
		}
		cout<<dp[k][n]<<endl;
	}
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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<cstring>
#define pb push_back
#include<vector>
using namespace std;
int n,k;
long long int dp[20][2000];
vector<int> p;
int main(){
	vector<bool> used(1130,true);
	for(int i=2;i<=1120;i++){
		if(used[i]){
			for(int j=i*2;j<=1120;j+=i){
				used[j]=false;
			}
			p.pb(i);
		}
	}
	while(cin>>n>>k,n||k){
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(int i=0;i<p.size()&&p[i]<=n;i++){//i番目の素数
			for(int j=k-1;j>=0;j--){//j回使った
				for(int l=0;l<=n-p[i];l++){//l
					if(dp[j][l]){
						dp[j+1][l+p[i]]+=dp[j][l];
					}
				}
			}
		}
		cout<<dp[k][n]<<endl;
	}
	return 0;
}
```