#include <stdio.h>
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;

#define MAXN 1121
#define MAXK 15

int N, K;
int dp[MAXN][MAXK];
int i, j;
vector<int> prime;

void makeprime(){
    bool primebool[MAXN];
    for(i = 0; i < MAXN; i++){
        primebool[i] = true;
    }

    primebool[0] = false;
    primebool[1] = false;

    for(i = 0; i < sqrt(MAXN); i++){
        if(!primebool[i]) continue;
        for(int j = 2 * i; j < MAXN; j += i){
            primebool[j] = false;
        }
    }

    for(i = 0; i < MAXN; i++){
        if(primebool[i]) prime.push_back(i);
    }
}


int main(){
    makeprime();
    while(true){
        cin >> N >> K;
        if(N == 0 && K == 0) break;

        for(i = 0; i < MAXN; i++){
            for(j = 0; j < MAXK; j++){
                dp[i][j] = 0;
            }
        }
        dp[0][0] = 1;

        for(int x: prime){
            if(x > N) break;
            for(j = K; j >= 0; j--){
                for(i = 0; i <= N; i++){
                    if(dp[i][j] != 0 && i + x <= N) dp[i + x][j + 1] += dp[i][j];
                }
            }
        }
        cout << dp[N][K] << endl;
    }
}


### Prompt

Your challenge is to write a CPP solution to 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 <stdio.h>
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;

#define MAXN 1121
#define MAXK 15

int N, K;
int dp[MAXN][MAXK];
int i, j;
vector<int> prime;

void makeprime(){
    bool primebool[MAXN];
    for(i = 0; i < MAXN; i++){
        primebool[i] = true;
    }

    primebool[0] = false;
    primebool[1] = false;

    for(i = 0; i < sqrt(MAXN); i++){
        if(!primebool[i]) continue;
        for(int j = 2 * i; j < MAXN; j += i){
            primebool[j] = false;
        }
    }

    for(i = 0; i < MAXN; i++){
        if(primebool[i]) prime.push_back(i);
    }
}


int main(){
    makeprime();
    while(true){
        cin >> N >> K;
        if(N == 0 && K == 0) break;

        for(i = 0; i < MAXN; i++){
            for(j = 0; j < MAXK; j++){
                dp[i][j] = 0;
            }
        }
        dp[0][0] = 1;

        for(int x: prime){
            if(x > N) break;
            for(j = K; j >= 0; j--){
                for(i = 0; i <= N; i++){
                    if(dp[i][j] != 0 && i + x <= N) dp[i + x][j + 1] += dp[i][j];
                }
            }
        }
        cout << dp[N][K] << endl;
    }
}


```