#include<iostream>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<sstream>
#include<iomanip>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef pair<int,pii> pip;
typedef vector<pip>vip;
const double PI=acos(-1);
const double EPS=1e-8;
const int inf=1e8;
int main(){
	int sosu[1150]={1,1,0};
	for(int i=2;i<1150;i++)if(sosu[i]==0)
	for(int j=i*2;j<1150;j+=i)sosu[j]=true;
	
	vi so;
	rep(i,1150)if(sosu[i]==0)so.pb(i);
	int dp[200][1150]={0};
	dp[0][0]=1;
	rep(i,so.size())for(int j=198;j>=0;j--)rep(k,1150)if(dp[j][k]){
		if(k+so[i]<1150)dp[j+1][k+so[i]]+=dp[j][k];
	}
	
	int n,k;
	while(cin>>n>>k,n+k){
		cout<<dp[k][n]<<endl;
	}
}
### 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>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<sstream>
#include<iomanip>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef pair<int,pii> pip;
typedef vector<pip>vip;
const double PI=acos(-1);
const double EPS=1e-8;
const int inf=1e8;
int main(){
	int sosu[1150]={1,1,0};
	for(int i=2;i<1150;i++)if(sosu[i]==0)
	for(int j=i*2;j<1150;j+=i)sosu[j]=true;
	
	vi so;
	rep(i,1150)if(sosu[i]==0)so.pb(i);
	int dp[200][1150]={0};
	dp[0][0]=1;
	rep(i,so.size())for(int j=198;j>=0;j--)rep(k,1150)if(dp[j][k]){
		if(k+so[i]<1150)dp[j+1][k+so[i]]+=dp[j][k];
	}
	
	int n,k;
	while(cin>>n>>k,n+k){
		cout<<dp[k][n]<<endl;
	}
}
```