#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (bit & (1LL << i))

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<typename T> using VVVV = std::vector<std::vector<std::vector<std::vector<T>>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<class T> inline std::vector<T> unique(std::vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    return v;
}

const int INF = 1 << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = { 0, 1, 0, -1 }, dx4 = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;

// nまでの素数のリストを作成(nを含む) O(N log log N)
vector<int> make_prime_list(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = false;
    is_prime[1] = false;

    vector<int> prime_list;
    for (int i = 2; i < n + 1; ++i) {
        if (!is_prime[i]) { continue; }
        prime_list.emplace_back(i);

        for (int j = i + i; j < n + 1; j += i) {
            is_prime[j] = false;
        }
    }
    return prime_list;  // 素数のリスト
    //return is_prime;  // すべての数値について素数かどうか
}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    V<int> prime_list = make_prime_list(1200);

    while (true) {
        int N, K;
        cin >> N >> K;
        if (N + K == 0) {
            break;
        }

        VVV<LL> dp(prime_list.size() + 1, VV<LL>(N + 1, V<LL>(K + 1, 0)));
        dp[0][0][0] = 1;
        FOR(i, 0, prime_list.size()) {
            FOR(j, 0, N + 1) {
                FOR(k, 0, K + 1) {

                    // not use
                    dp[i + 1][j][k] += dp[i][j][k];

                    // use
                    if (j + prime_list[i] <= N and k + 1 <= K) {
                        dp[i + 1][j + prime_list[i]][k + 1] += dp[i][j][k];
                    }
                }
            }
        }
        print(dp[prime_list.size()][N][K]);
    }

    return 0;
}

### Prompt

Develop a solution in CPP 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 <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (bit & (1LL << i))

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<typename T> using VVVV = std::vector<std::vector<std::vector<std::vector<T>>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<class T> inline std::vector<T> unique(std::vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    return v;
}

const int INF = 1 << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = { 0, 1, 0, -1 }, dx4 = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;

// nまでの素数のリストを作成(nを含む) O(N log log N)
vector<int> make_prime_list(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = false;
    is_prime[1] = false;

    vector<int> prime_list;
    for (int i = 2; i < n + 1; ++i) {
        if (!is_prime[i]) { continue; }
        prime_list.emplace_back(i);

        for (int j = i + i; j < n + 1; j += i) {
            is_prime[j] = false;
        }
    }
    return prime_list;  // 素数のリスト
    //return is_prime;  // すべての数値について素数かどうか
}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    V<int> prime_list = make_prime_list(1200);

    while (true) {
        int N, K;
        cin >> N >> K;
        if (N + K == 0) {
            break;
        }

        VVV<LL> dp(prime_list.size() + 1, VV<LL>(N + 1, V<LL>(K + 1, 0)));
        dp[0][0][0] = 1;
        FOR(i, 0, prime_list.size()) {
            FOR(j, 0, N + 1) {
                FOR(k, 0, K + 1) {

                    // not use
                    dp[i + 1][j][k] += dp[i][j][k];

                    // use
                    if (j + prime_list[i] <= N and k + 1 <= K) {
                        dp[i + 1][j + prime_list[i]][k + 1] += dp[i][j][k];
                    }
                }
            }
        }
        print(dp[prime_list.size()][N][K]);
    }

    return 0;
}

```