#include <bits/stdc++.h>
using namespace std;

using ll = long long int;

constexpr int MAX_N = 1125;

vector<bool> IsPrime;
vector<int> primes;

void sieve(int max) {
    IsPrime.resize(max + 1, true);
    IsPrime[0] = IsPrime[1] = false;
    for (int i = 2; i * i <= max; ++i){
        if (IsPrime[i]){
            for (int j = i * i; j <= max; j += i) IsPrime[j] = false;
        }
    }
    for(int i = 0; i <= max; ++i) if(IsPrime[i]) primes.push_back(i);
}

ll memo[MAX_N][200][15];

ll rec(int n, int m, int k){
    if(k == 0) return n == 0;
    if(m < 0) return 0;
    if(memo[n][m][k] >= 0) return memo[n][m][k];
    ll res = 0;
    for(int i=m;i>=0;i--){
        if(primes[i] <= n){
            res += rec(n-primes[i],i-1,k-1);
        }
    }
    return memo[n][m][k] = res;
}

int solve(){
    int n, k;
    cin >> n >> k;
    if(n == 0) return 1;
    cout << rec(n, int(primes.size())-1, k) << endl;
    return 0;
}

int main(){
    sieve(MAX_N);
    for(int i=0;i<MAX_N;i++){
        for(int j=0;j<200;j++){
            fill(memo[i][j],memo[i][j]+15,-1LL);
        }
    }
    while(solve()==0);
    return 0;
}


### Prompt

Your task is to create 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 <bits/stdc++.h>
using namespace std;

using ll = long long int;

constexpr int MAX_N = 1125;

vector<bool> IsPrime;
vector<int> primes;

void sieve(int max) {
    IsPrime.resize(max + 1, true);
    IsPrime[0] = IsPrime[1] = false;
    for (int i = 2; i * i <= max; ++i){
        if (IsPrime[i]){
            for (int j = i * i; j <= max; j += i) IsPrime[j] = false;
        }
    }
    for(int i = 0; i <= max; ++i) if(IsPrime[i]) primes.push_back(i);
}

ll memo[MAX_N][200][15];

ll rec(int n, int m, int k){
    if(k == 0) return n == 0;
    if(m < 0) return 0;
    if(memo[n][m][k] >= 0) return memo[n][m][k];
    ll res = 0;
    for(int i=m;i>=0;i--){
        if(primes[i] <= n){
            res += rec(n-primes[i],i-1,k-1);
        }
    }
    return memo[n][m][k] = res;
}

int solve(){
    int n, k;
    cin >> n >> k;
    if(n == 0) return 1;
    cout << rec(n, int(primes.size())-1, k) << endl;
    return 0;
}

int main(){
    sieve(MAX_N);
    for(int i=0;i<MAX_N;i++){
        for(int j=0;j<200;j++){
            fill(memo[i][j],memo[i][j]+15,-1LL);
        }
    }
    while(solve()==0);
    return 0;
}


```