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>
using namespace std;

const int MAX_N = 1120;
const int MAX_K = 14;

// is_p[i] := i が素数のとき true
bool is_p[MAX_N+1] = {false,false};
int dp[MAX_K+1][MAX_N+1] = {0};

// エラトステネスのふるい
void sieve(){
	for(int i = 2 ; i <= MAX_N ; i++ )
		is_p[i] = true;
	for(int i = 2 ; i*i <= MAX_N ; i++ ){
		if( is_p[i] ){
			for(int j = i+i ; j <= MAX_N ; j += i ){
				is_p[j] = false;
			}
		}
	}
}

int main(){
	sieve();
	dp[0][0] = 1;
	for(int i=2 ; i <= MAX_N ; i++ ){
		if( is_p[i] ){
			for(int j = MAX_K; j > 0; j-- ){
				for(int k = i ; k <= MAX_N ; k++ ){
					dp[j][k] += dp[j-1][k-i];
				}
			}
		}
	}
	int n, k;
	while( cin >> n >> k , n || k ){
		cout << dp[k][n] << endl;
	}
}