#include<stdio.h>
int isp[1121];
int prime[500];
int a,b;
int dp[1121][15];
int now;
int main(){
	isp[0]=isp[1]=-1;
	for(int i=2;i<=1120;i++){
		if(isp[i]==0){
			isp[i]=1;
			for(int j=i*2;j<=1120;j+=i)isp[j]=-1;
		}
	}
	for(int i=0;i<1121;i++)if(~isp[i])prime[now++]=i;
	dp[0][0]=1;
	for(int i=0;i<now;i++){
			for(int j=13;j>=0;j--){
		for(int k=0;k<1121;k++){
				if(k+prime[i]<1121)dp[k+prime[i]][j+1]+=dp[k][j];
			}
		}
	}
	while(scanf("%d%d",&a,&b),a+b){
		printf("%d\n",dp[a][b]);
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include<stdio.h>
int isp[1121];
int prime[500];
int a,b;
int dp[1121][15];
int now;
int main(){
	isp[0]=isp[1]=-1;
	for(int i=2;i<=1120;i++){
		if(isp[i]==0){
			isp[i]=1;
			for(int j=i*2;j<=1120;j+=i)isp[j]=-1;
		}
	}
	for(int i=0;i<1121;i++)if(~isp[i])prime[now++]=i;
	dp[0][0]=1;
	for(int i=0;i<now;i++){
			for(int j=13;j>=0;j--){
		for(int k=0;k<1121;k++){
				if(k+prime[i]<1121)dp[k+prime[i]][j+1]+=dp[k][j];
			}
		}
	}
	while(scanf("%d%d",&a,&b),a+b){
		printf("%d\n",dp[a][b]);
	}
}
```