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

const int MAX = 1120;
int primes[200];
int primesSize;
int c[1121][15][188];

int solve(int n, int k, int index) {
    if (n == 0 && k == 0) {
        return 1;
    } else if (n == 0 || k == 0) {
        return 0;
    }
    if (c[n][k][index] >= 0) {
        return c[n][k][index];
    }
    int res = 0;
    for (int i = index; i < primesSize; i++) {
        if (primes[i] > n) {
            break;
        } else {
            res += solve(n - primes[i], k - 1, i + 1);
        }
    }
    c[n][k][index] = res;
    return res;
}

int main() {
    primes[0] = 2;
    primesSize = 1;
    for (int i = 3; i <= MAX; i += 2) {
        bool add = true;
        for (int j = 0; j < primesSize; j++) {
            if (primes[j] * primes[j] > i) {
                break;
            }
            if (i % primes[j] == 0) {
                add = false;
                break;
            }
        }
        if (add) {
            primes[primesSize++] = i;
        }
    }
    memset(c, -1, sizeof(c));
    while (true) {
        int n, k;
        cin >> n >> k;
        if ((n | k) == 0) {
            break;
        }
        cout << solve(n, k, 0) << endl;
    }
}
### Prompt

In CPP, your task is to solve 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 <cstring>
using namespace std;

const int MAX = 1120;
int primes[200];
int primesSize;
int c[1121][15][188];

int solve(int n, int k, int index) {
    if (n == 0 && k == 0) {
        return 1;
    } else if (n == 0 || k == 0) {
        return 0;
    }
    if (c[n][k][index] >= 0) {
        return c[n][k][index];
    }
    int res = 0;
    for (int i = index; i < primesSize; i++) {
        if (primes[i] > n) {
            break;
        } else {
            res += solve(n - primes[i], k - 1, i + 1);
        }
    }
    c[n][k][index] = res;
    return res;
}

int main() {
    primes[0] = 2;
    primesSize = 1;
    for (int i = 3; i <= MAX; i += 2) {
        bool add = true;
        for (int j = 0; j < primesSize; j++) {
            if (primes[j] * primes[j] > i) {
                break;
            }
            if (i % primes[j] == 0) {
                add = false;
                break;
            }
        }
        if (add) {
            primes[primesSize++] = i;
        }
    }
    memset(c, -1, sizeof(c));
    while (true) {
        int n, k;
        cin >> n >> k;
        if ((n | k) == 0) {
            break;
        }
        cout << solve(n, k, 0) << endl;
    }
}
```