#include<algorithm>
#include<bitset>
#include<cassert>
#include<cfloat>
#include<climits>
#include<cmath>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define REVERSE(x) reverse((x).begin(), (x).end())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << '\n'
#define endl '\n'

constexpr int IINF = (1 << 30) - 1;
constexpr long long LLINF = 1LL << 61;
constexpr double EPS = 1e-10;

constexpr int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
constexpr int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

template<typename T>
bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T& a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

std::vector<int> primeNumbers;
std::vector<bool> isPrime;

void eratosthenes(int n){
    int i, j;
    int s = 4;

    primeNumbers.clear();
    primeNumbers = {2, 3};

    isPrime.assign(std::max(n + 1, 2), true);
    isPrime[0] = false;
    isPrime[1] = false;

    for(i = 4 ; i <= n ; i += 2){
        isPrime[i] = false;
    }
    for(i = 9 ; i <= n ; i += 6){
        isPrime[i] = false;
    }

    for(i = 5 ; i * i <= n ; i += s){
        if(isPrime[i]){
            primeNumbers.emplace_back(i);
            for(j = i * i ; j <= n ; j += i){
                isPrime[j] = false;
            }
        }
        s ^= 6;
    }
    for( ; i <= n ; i += s){
        if(isPrime[i]){
            primeNumbers.emplace_back(i);
        }
        s ^= 6;
    }

    return;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    constexpr int N = 1120;
    constexpr int K = 14;

    eratosthenes(N + 1);

    const int m = (int)primeNumbers.size();

    vector< vector< vector<int> > > dp(m + 1, vector< vector<int> >(N + 1, vector<int>(K + 1, 0)));
    dp[0][0][0] = 1;

    rep(i, m){
        rep(j, N + 1){
            rep(l, K + 1){
                dp[i + 1][j][l] += dp[i][j][l];
                if(j + primeNumbers[i] <= N && l + 1 <= K){
                    dp[i + 1][j + primeNumbers[i]][l + 1] += dp[i][j][l];
                }
            }
        }
    }

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

    cout << flush;

    return 0;
}

### Prompt

Create a solution in Cpp for 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<cassert>
#include<cfloat>
#include<climits>
#include<cmath>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define REVERSE(x) reverse((x).begin(), (x).end())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << '\n'
#define endl '\n'

constexpr int IINF = (1 << 30) - 1;
constexpr long long LLINF = 1LL << 61;
constexpr double EPS = 1e-10;

constexpr int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
constexpr int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

template<typename T>
bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T& a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

std::vector<int> primeNumbers;
std::vector<bool> isPrime;

void eratosthenes(int n){
    int i, j;
    int s = 4;

    primeNumbers.clear();
    primeNumbers = {2, 3};

    isPrime.assign(std::max(n + 1, 2), true);
    isPrime[0] = false;
    isPrime[1] = false;

    for(i = 4 ; i <= n ; i += 2){
        isPrime[i] = false;
    }
    for(i = 9 ; i <= n ; i += 6){
        isPrime[i] = false;
    }

    for(i = 5 ; i * i <= n ; i += s){
        if(isPrime[i]){
            primeNumbers.emplace_back(i);
            for(j = i * i ; j <= n ; j += i){
                isPrime[j] = false;
            }
        }
        s ^= 6;
    }
    for( ; i <= n ; i += s){
        if(isPrime[i]){
            primeNumbers.emplace_back(i);
        }
        s ^= 6;
    }

    return;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    constexpr int N = 1120;
    constexpr int K = 14;

    eratosthenes(N + 1);

    const int m = (int)primeNumbers.size();

    vector< vector< vector<int> > > dp(m + 1, vector< vector<int> >(N + 1, vector<int>(K + 1, 0)));
    dp[0][0][0] = 1;

    rep(i, m){
        rep(j, N + 1){
            rep(l, K + 1){
                dp[i + 1][j][l] += dp[i][j][l];
                if(j + primeNumbers[i] <= N && l + 1 <= K){
                    dp[i + 1][j + primeNumbers[i]][l + 1] += dp[i][j][l];
                }
            }
        }
    }

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

    cout << flush;

    return 0;
}

```