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

#define REP(i,n,m) for(int i=n;i<m;i++)
#define rep(i,n) REP(i,0,n)

int dp[1130][15][200];
bool flg[1130];
vector<int> p;

void mkPrime(void){
	memset(flg,1,sizeof(flg));
	flg[0] = flg[1] = 0;

	REP(i,2,1121){
		if(flg[i]){
			for(int j=i*2;j<=1120;j+=i){
				flg[j] = false;
			}
		}
	}
}

int rec(int n,int k,int idx){
	if(n == 0 && k == 0) return 1;
	if(n < 0 || k < 0) return 0;
	if(dp[n][k][idx]!=-1) return dp[n][k][idx];

	int res = 0;
	for(int i=idx;p[i]<=n;i++){
		res += rec(n-p[i],k-1,i+1);
	}
	return dp[n][k][idx]=res;
}

int main(void){
	memset(dp,-1,sizeof(dp));
	mkPrime();
	rep(i,1130) if(flg[i]) p.push_back(i);

	int n,k;
	while(cin>>n>>k && (n||k)){
		cout<<rec(n,k,0)<<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>
#include <vector>
using namespace std;

#define REP(i,n,m) for(int i=n;i<m;i++)
#define rep(i,n) REP(i,0,n)

int dp[1130][15][200];
bool flg[1130];
vector<int> p;

void mkPrime(void){
	memset(flg,1,sizeof(flg));
	flg[0] = flg[1] = 0;

	REP(i,2,1121){
		if(flg[i]){
			for(int j=i*2;j<=1120;j+=i){
				flg[j] = false;
			}
		}
	}
}

int rec(int n,int k,int idx){
	if(n == 0 && k == 0) return 1;
	if(n < 0 || k < 0) return 0;
	if(dp[n][k][idx]!=-1) return dp[n][k][idx];

	int res = 0;
	for(int i=idx;p[i]<=n;i++){
		res += rec(n-p[i],k-1,i+1);
	}
	return dp[n][k][idx]=res;
}

int main(void){
	memset(dp,-1,sizeof(dp));
	mkPrime();
	rep(i,1130) if(flg[i]) p.push_back(i);

	int n,k;
	while(cin>>n>>k && (n||k)){
		cout<<rec(n,k,0)<<endl;
	}

	return 0;
}
```