#include <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define repi(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, a) repi(i, 0, a)
#define repd(i, a, b) for(int i = (a); i >= (b); i--)
#define repit(i, a) for(__typeof((a).begin()) i = (a).begin(); i != (a).end(); i++)

#define all(u) (u).begin(), (u).end()
#define rall(u) (u).rbegin(), (u).rend()
#define UNIQUE(u) (u).erase(unique(all(u)), (u).end())

#define pb push_back
#define mp make_pair
#define INF 1e9
#define EPS 1e-9
#define PI acos(-1.0)

using namespace std;

typedef long long ll;
typedef vector<int> vi;

const int N = 1200;

int dp[N][16] = {1}, c[N] = {1, 1};

void gen()
{
    for (int i = 2; i < N; ++i) {
	if (!c[i]) {
	    for (int j = N - 2; j >= i; --j) {
		for (int k = 1; k < 16; ++k) {
		    dp[j][k] += dp[j - i][k - 1];
		}
	    }
	    for (int j = i + i; j < N; j += i) {
		c[j] = 1;
	    }
	}
    }
}

int main(){
    gen();

    int n, k;
    while (cin >> n >> k && n) {
	cout << dp[n][k] << 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 <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define repi(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, a) repi(i, 0, a)
#define repd(i, a, b) for(int i = (a); i >= (b); i--)
#define repit(i, a) for(__typeof((a).begin()) i = (a).begin(); i != (a).end(); i++)

#define all(u) (u).begin(), (u).end()
#define rall(u) (u).rbegin(), (u).rend()
#define UNIQUE(u) (u).erase(unique(all(u)), (u).end())

#define pb push_back
#define mp make_pair
#define INF 1e9
#define EPS 1e-9
#define PI acos(-1.0)

using namespace std;

typedef long long ll;
typedef vector<int> vi;

const int N = 1200;

int dp[N][16] = {1}, c[N] = {1, 1};

void gen()
{
    for (int i = 2; i < N; ++i) {
	if (!c[i]) {
	    for (int j = N - 2; j >= i; --j) {
		for (int k = 1; k < 16; ++k) {
		    dp[j][k] += dp[j - i][k - 1];
		}
	    }
	    for (int j = i + i; j < N; j += i) {
		c[j] = 1;
	    }
	}
    }
}

int main(){
    gen();

    int n, k;
    while (cin >> n >> k && n) {
	cout << dp[n][k] << endl;
    }

    return 0;
}
```