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

#define MAX_N 1120+1
#define MAX_K 14+1
#define MAX_L 187+1

vector<int> p;
int dp[MAX_N][MAX_K][MAX_L];

void init() {
    memset(dp, -1, sizeof(dp));
    bool f[MAX_N];
    memset(f, 1, sizeof(f));
    f[0] = f[1] = false;
    for (int i = 2; i <= MAX_N; i++) 
        if (f[i]) {
            p.push_back(i);
            for (int j = i+i; j <= MAX_N; j+=i)
                f[j] = false;
        }
}

int F(int n, int k, int l) {
    //cout << n << ' ' << k << ' ' << l << endl;
    if (dp[n][k][l] >= 0) return dp[n][k][l];
    if (k == 0) {
        if (n == 0) return 1;
        return 0;
    }
    if (n < 2) return 0;
    int c = 0;
    for (int i = 0; i < p.size() && p[i] <= n && i < l; i++) {
        c += F(n-p[i], k-1, i);
    }
    return dp[n][k][l] = c;
}

int main() {
    init();
    int n, k;
    while (cin >> n >> k) {
        if (n == 0 && k == 0) break;
        cout << F(n, k, p.size()) << endl;
    }
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#include <cstring>
using namespace std;

#define MAX_N 1120+1
#define MAX_K 14+1
#define MAX_L 187+1

vector<int> p;
int dp[MAX_N][MAX_K][MAX_L];

void init() {
    memset(dp, -1, sizeof(dp));
    bool f[MAX_N];
    memset(f, 1, sizeof(f));
    f[0] = f[1] = false;
    for (int i = 2; i <= MAX_N; i++) 
        if (f[i]) {
            p.push_back(i);
            for (int j = i+i; j <= MAX_N; j+=i)
                f[j] = false;
        }
}

int F(int n, int k, int l) {
    //cout << n << ' ' << k << ' ' << l << endl;
    if (dp[n][k][l] >= 0) return dp[n][k][l];
    if (k == 0) {
        if (n == 0) return 1;
        return 0;
    }
    if (n < 2) return 0;
    int c = 0;
    for (int i = 0; i < p.size() && p[i] <= n && i < l; i++) {
        c += F(n-p[i], k-1, i);
    }
    return dp[n][k][l] = c;
}

int main() {
    init();
    int n, k;
    while (cin >> n >> k) {
        if (n == 0 && k == 0) break;
        cout << F(n, k, p.size()) << endl;
    }
    return 0;
}
```