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

int p[1130];
int p_size;

void sieve(){
	p[0] = 2;
	p_size = 1;
	for(int i = 3 ; i <= 1120 ; i += 2 ){
		bool flag = true;
		for(int j = 0 ; j < p_size ; j++ ){
			if( i < p[j] * p[j] ){
				break;
			}else if( i % p[j] == 0 ){
				flag = false;
				break;
			}
		}
		if( flag ){
			p[p_size++] = i;
		}
	}
}

int dp[1130][15][200];

int solve(int n, int k, int last = 0){
	if( n == 0 && k == 0 ){
		return 1;
	}else if( n == 0 || k == 0 ){
		return 0;
	}else if( dp[n][k][last] != -1 ){
		return dp[n][k][last];
	}
	
	int res = 0;
	for(int i = last ; i < p_size ; i++ ){
		if( n < p[i] ){
			break;
		}else{
			res += solve(n-p[i], k-1, i+1);
		}
	}
	return dp[n][k][last] = res;
}

int main(){
	int n, k;
	sieve();
	for(int i=0 ; i < 1130 ; i++ )
		for(int j=0 ; j < 15 ; j++ )
			for(int k=0 ; k < 200 ; k++ )
				dp[i][j][k] = -1;
	
	while( cin >> n >> k , n || k ){
		cout << solve(n, k) << endl;
	}
}
### Prompt

Your task is to create a Cpp solution 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 <vector>
using namespace std;

int p[1130];
int p_size;

void sieve(){
	p[0] = 2;
	p_size = 1;
	for(int i = 3 ; i <= 1120 ; i += 2 ){
		bool flag = true;
		for(int j = 0 ; j < p_size ; j++ ){
			if( i < p[j] * p[j] ){
				break;
			}else if( i % p[j] == 0 ){
				flag = false;
				break;
			}
		}
		if( flag ){
			p[p_size++] = i;
		}
	}
}

int dp[1130][15][200];

int solve(int n, int k, int last = 0){
	if( n == 0 && k == 0 ){
		return 1;
	}else if( n == 0 || k == 0 ){
		return 0;
	}else if( dp[n][k][last] != -1 ){
		return dp[n][k][last];
	}
	
	int res = 0;
	for(int i = last ; i < p_size ; i++ ){
		if( n < p[i] ){
			break;
		}else{
			res += solve(n-p[i], k-1, i+1);
		}
	}
	return dp[n][k][last] = res;
}

int main(){
	int n, k;
	sieve();
	for(int i=0 ; i < 1130 ; i++ )
		for(int j=0 ; j < 15 ; j++ )
			for(int k=0 ; k < 200 ; k++ )
				dp[i][j][k] = -1;
	
	while( cin >> n >> k , n || k ){
		cout << solve(n, k) << endl;
	}
}
```