#include <iostream>
#include <array>
#include <vector>
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
#define repeat(i,n) repeat_from(i,0,n)
#define repeat_from_reverse(i,m,n) for (int i = (n)-1; (i) >= (m); --(i))
#define repeat_reverse(i,n) repeat_from_reverse(i,0,n)
typedef long long ll;
using namespace std;
vector<int> primes(int n) {
    vector<bool> is_prime(n+1, true);
    is_prime[0] = false;
    is_prime[1] = false;
    for (int i = 2; i * i <= n; ++ i) {
        if (is_prime[i]) {
            for (int j = 2*i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }
    vector<int> result;
    for (int i = 0; i <= n; ++ i) {
        if (is_prime[i]) {
            result.push_back(i);
        }
    }
    return result;
}
constexpr int max_n = 1120;
constexpr int max_k = 14;
int main() {
    vector<int> ps = primes(max_n);
    array<array<ll,max_k+1>,max_n+1> dp = {};
    dp[0][0] = 1;
    for (int p : ps) {
        repeat_reverse (k, max_k+1) {
            repeat (n, max_n+1) {
                if (n-p >= 0 and k-1 >= 0) {
                    dp[n][k] += dp[n-p][k-1];
                }
            }
        }
    }
    while (true) {
        int n, k; cin >> n >> k;
        if (n == 0 and k == 0) break;
        cout << dp[n][k] << endl;
    }
    return 0;
}
### 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 <array>
#include <vector>
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
#define repeat(i,n) repeat_from(i,0,n)
#define repeat_from_reverse(i,m,n) for (int i = (n)-1; (i) >= (m); --(i))
#define repeat_reverse(i,n) repeat_from_reverse(i,0,n)
typedef long long ll;
using namespace std;
vector<int> primes(int n) {
    vector<bool> is_prime(n+1, true);
    is_prime[0] = false;
    is_prime[1] = false;
    for (int i = 2; i * i <= n; ++ i) {
        if (is_prime[i]) {
            for (int j = 2*i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }
    vector<int> result;
    for (int i = 0; i <= n; ++ i) {
        if (is_prime[i]) {
            result.push_back(i);
        }
    }
    return result;
}
constexpr int max_n = 1120;
constexpr int max_k = 14;
int main() {
    vector<int> ps = primes(max_n);
    array<array<ll,max_k+1>,max_n+1> dp = {};
    dp[0][0] = 1;
    for (int p : ps) {
        repeat_reverse (k, max_k+1) {
            repeat (n, max_n+1) {
                if (n-p >= 0 and k-1 >= 0) {
                    dp[n][k] += dp[n-p][k-1];
                }
            }
        }
    }
    while (true) {
        int n, k; cin >> n >> k;
        if (n == 0 and k == 0) break;
        cout << dp[n][k] << endl;
    }
    return 0;
}
```