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<bits/stdc++.h>
#define N 1150
using namespace std;
bool prime[N];
vector<int> p;
int dp[205][N][15];

int main(){
  for(int i=0;i<N;i++)
    prime[i]=true;
  prime[0]=prime[1]=false;
  for(int i=0;i*i<N;i++){
    if(!prime[i])continue;
    for(int j=i*2;j<N;j+=i)
      prime[j]=false;
  }
  for(int i=0;i<N;i++)
    if(prime[i])p.push_back(i);
  while(1){
    int n,k;
    cin>>n>>k;
    if(!n&&!k)break;
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1;
    for(int i=0;i<p.size();i++){
      for(int j=0;j<=n;j++){
	for(int l=0;l<=k;l++){
	  if(!dp[i][j][l])continue;
	  if(j+p[i]<=n)
	    dp[i+1][j+p[i]][l+1]+=dp[i][j][l];
	  dp[i+1][j][l]+=dp[i][j][l];
	}
      }
    }
    cout<<dp[p.size()][n][k]<<endl;
  }
  return 0;
}