#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>

using namespace std;

long long dp[200][15][1121];
bool isprime[1121];
vector<int> primes;

int main(){
    memset(isprime, true, sizeof(isprime));
    for(int i=2;i<=1120;++i){
        if(isprime[i]){
            primes.push_back(i);
            for(int j=2;i*j<=1120;++j){
                isprime[i*j] = false;
            }
        }
    }
    int sz = primes.size();

    dp[0][0][0] = 1;
    
    for(int i=1;i<=sz;i++){
        for(int j=0;j<=14;j++){
            for(int k=0;k<=1120;k++){
                if(k+primes[i-1] <= 1120 && j != 14){
                    dp[i][j+1][k+primes[i-1]] += dp[i-1][j][k];
                }
                dp[i][j][k] += dp[i-1][j][k];
            }
        }
    }

    int n,k;
    while(cin >> n >> k){
        if((n|k) == 0) break;
        cout << dp[sz][k][n] << endl;
    }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <algorithm>
#include <vector>
#include <cstring>

using namespace std;

long long dp[200][15][1121];
bool isprime[1121];
vector<int> primes;

int main(){
    memset(isprime, true, sizeof(isprime));
    for(int i=2;i<=1120;++i){
        if(isprime[i]){
            primes.push_back(i);
            for(int j=2;i*j<=1120;++j){
                isprime[i*j] = false;
            }
        }
    }
    int sz = primes.size();

    dp[0][0][0] = 1;
    
    for(int i=1;i<=sz;i++){
        for(int j=0;j<=14;j++){
            for(int k=0;k<=1120;k++){
                if(k+primes[i-1] <= 1120 && j != 14){
                    dp[i][j+1][k+primes[i-1]] += dp[i-1][j][k];
                }
                dp[i][j][k] += dp[i-1][j][k];
            }
        }
    }

    int n,k;
    while(cin >> n >> k){
        if((n|k) == 0) break;
        cout << dp[sz][k][n] << endl;
    }
}
```