#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;
const int MAX_PRIME=2000;
int prime[MAX_PRIME+1];
bool isPrime[MAX_PRIME+1];
int p;
void erats(){
	// primeツδ環スツトツづーツ渉可甘コツ可サ
	fill(isPrime,isPrime+MAX_PRIME,true);
	p = 0;
	isPrime[0] = isPrime[1] = false;

	for(int i = 2; i <= MAX_PRIME; i++){
		if(isPrime[i]){
			prime[p++] = i;
		}
		else
			continue;

		// ツづ督づゥツつ「ツづ可つゥツつッツづゥ
		for(int j = 2*i; j <= MAX_PRIME; j+=i){
			isPrime[j] = false;
		}
		cout << flush;
	}
}

int n,k;
int dp[15][1200][305];
int dfs(int cnt,int left,int tail){
    if(dp[cnt][left][tail]!=-1)return dp[cnt][left][tail];
    int res=0;
    if(cnt==k){
        if(left==0)
            return 1;
        else
            return 0;
    }
    for(int i = tail; prime[i] <= left; i++){
        res+=dfs(cnt+1,left-prime[i],i+1);
    }
    return dp[cnt][left][tail]=res;
}

int main(){
    erats();
    while(cin>>n>>k&&!(n==0&&k==0)){
        memset(dp,-1,sizeof(dp));
        int res=dfs(0,n,0);
        cout<<res<<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 <algorithm>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;
const int MAX_PRIME=2000;
int prime[MAX_PRIME+1];
bool isPrime[MAX_PRIME+1];
int p;
void erats(){
	// primeツδ環スツトツづーツ渉可甘コツ可サ
	fill(isPrime,isPrime+MAX_PRIME,true);
	p = 0;
	isPrime[0] = isPrime[1] = false;

	for(int i = 2; i <= MAX_PRIME; i++){
		if(isPrime[i]){
			prime[p++] = i;
		}
		else
			continue;

		// ツづ督づゥツつ「ツづ可つゥツつッツづゥ
		for(int j = 2*i; j <= MAX_PRIME; j+=i){
			isPrime[j] = false;
		}
		cout << flush;
	}
}

int n,k;
int dp[15][1200][305];
int dfs(int cnt,int left,int tail){
    if(dp[cnt][left][tail]!=-1)return dp[cnt][left][tail];
    int res=0;
    if(cnt==k){
        if(left==0)
            return 1;
        else
            return 0;
    }
    for(int i = tail; prime[i] <= left; i++){
        res+=dfs(cnt+1,left-prime[i],i+1);
    }
    return dp[cnt][left][tail]=res;
}

int main(){
    erats();
    while(cin>>n>>k&&!(n==0&&k==0)){
        memset(dp,-1,sizeof(dp));
        int res=dfs(0,n,0);
        cout<<res<<endl;
    }
    return 0;
}
```