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
import java.util.*;
import java.io.*;
 
public class Main{
    public static void main(String[] args){
        new Main().solve();
    }
 
    void solve(){
 
        Scanner cin = new Scanner(System.in);
        int MAXA = 1200;
        int MAXB = 20;
        int[][] dp = new int[MAXA][MAXB];
        dp[0][0] = 1;
        for(int i=2;i<=MAXA;i++){
            if(isPrime(i)){
                for(int j=MAXA- i - 1;j>= 0;j--){
                	for(int k=MAXB-2;k>=0;k--){
                		dp[j+i][k+1] += dp[j][k];
                	}
                }
            }
        }
        while(true)
        {
            int n = cin.nextInt();
            int k = cin.nextInt();
            if(n==0) return;
            System.out.println(dp[n][k]);
        }
    }
 
    boolean isPrime(int a){
        if(a<2) return false;
        for(int i=2;i*i<=a;i++){
            if(a%i==0) return false;
        }
        return true;
    }
}