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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#define INF (2000000000)
#define MAX (2000)
using namespace std;

typedef vector<int> VI;
typedef vector<int>::iterator VII;
typedef vector<double> VD;
typedef vector<double>::iterator VDI;

bool prime[MAX + 1];
int dp[15][1121];

int main()
{
	for (int i = 2; i <= MAX; i++){
		prime[i] = true;
	}
	for (int i = 2; i * i <= MAX; i++){
		for (int j = i * i; j <= MAX; j += i){
			prime[j] = false;
		}
	}
	
	dp[0][0] = 1;
	for (int i = 0; i <= 1120; i++){
		if (!prime[i]) continue;
		
		for (int j = 14; j >= 1; j--){
			for (int k = i; k <= 1120; k++){
				dp[j][k] += dp[j - 1][k - i];
			}
		}
	}
	
	int n, k;
	while (scanf("%d %d", &n, &k), n + k){
		printf("%d\n", dp[k][n]);
	}
}