#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>

using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype
template <class T> int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }
#endif

#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define popcount __builtin_popcount

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

template <class T> void max_swap(T& a, const T& b) { a = max(a, b); }
template <class T> void min_swap(T& a, const T& b) { a = min(a, b); }

typedef long long ll;
typedef pair<int, int> pint;

const double PI = acos(-1.0);
const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };



const int MAX_PRIME = 1200;
bool is_prime[MAX_PRIME + 1];
vector<int> primes;
void sieve()
{
	fill_n(is_prime, MAX_PRIME + 1, true);
	is_prime[0] = is_prime[1] = false;
	for (int i = 2; i * i <= MAX_PRIME; ++i)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= MAX_PRIME; j += i)
				is_prime[j] = false;
		}
	}
	for (int i = 0; i <= MAX_PRIME; ++i)
		if (is_prime[i])
			primes.push_back(i);
}
int main()
{
	sieve();

	static int dp[16][1200];
	dp[0][0] = 1;
	for (int i = 0; primes[i] <= 1120; ++i)
		for (int j = 14; j > 0; --j)
			for (int k = primes[i]; k <= 1120; ++k)
				dp[j][k] += dp[j - 1][k - primes[i]];

	int n, k;
	while (scanf("%d%d", &n, &k), n | k)
		printf("%d\n", dp[k][n]);
}
### 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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>

using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype
template <class T> int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }
#endif

#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define popcount __builtin_popcount

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

template <class T> void max_swap(T& a, const T& b) { a = max(a, b); }
template <class T> void min_swap(T& a, const T& b) { a = min(a, b); }

typedef long long ll;
typedef pair<int, int> pint;

const double PI = acos(-1.0);
const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };



const int MAX_PRIME = 1200;
bool is_prime[MAX_PRIME + 1];
vector<int> primes;
void sieve()
{
	fill_n(is_prime, MAX_PRIME + 1, true);
	is_prime[0] = is_prime[1] = false;
	for (int i = 2; i * i <= MAX_PRIME; ++i)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= MAX_PRIME; j += i)
				is_prime[j] = false;
		}
	}
	for (int i = 0; i <= MAX_PRIME; ++i)
		if (is_prime[i])
			primes.push_back(i);
}
int main()
{
	sieve();

	static int dp[16][1200];
	dp[0][0] = 1;
	for (int i = 0; primes[i] <= 1120; ++i)
		for (int j = 14; j > 0; --j)
			for (int k = primes[i]; k <= 1120; ++k)
				dp[j][k] += dp[j - 1][k - primes[i]];

	int n, k;
	while (scanf("%d%d", &n, &k), n | k)
		printf("%d\n", dp[k][n]);
}
```