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
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;

inline bool isPrime(int x){
  for(int i=2;i*i<=x;i++){
    if(x%i == 0)return false;
  }
  return true;
}

int n,k;
vector<int> prime;
int memo[1200][20];

int main(){
  for(int i=2;i<=1120;i++){
    if(isPrime(i))prime.push_back(i);
  }
  while(cin >> n >> k,n){
    memset(memo,0,sizeof(memo));
    memo[0][0] = 1;

    for(int x=0;x<prime.size();x++){
      if(prime[x] > n)break;

      for(int j=k-1;j>=0;j--){
	for(int i=n-prime[x];i>=0;i--){
	  memo[i+prime[x]][j+1] += memo[i][j];
	}
      }
    }

    cout << memo[n][k] << endl;
  }
}