#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define rep(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define all(v) (v).begin(), (v).end()
#define rev(s) (s).rbegin(), (s).rend()
#define MP make_pair
#define X first
#define Y second

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int INF = 100000000;

bool isprime[1500];

vector<int> prime;

void setprime(){
	fill(isprime, isprime+1500, true);
	isprime[0] = 	isprime[1] = false;
	for(int i = 2; i < 1500; i++ ){
		if(isprime[i]){
			prime.push_back(i);
			for(int j = i*2; j < 1500; j += i){
				isprime[j] = false;
			}
		}
	}
}

int dp[15][1200] = {};
void setdp(){
	setprime();

	for(int i = 0; i < prime.size(); i++){ // i番目までの素数を使う
		for(int j = 14; j >= 1; j--){	// j個の数字を使う
			if(j == 1) dp[1][prime[i]] += 1;
			for(int k = 0; k+prime[i] < 1200; k++){
				if(!dp[j-1][k]) continue;
				dp[j][k+prime[i]] += dp[j-1][k];
			}
		}
	}
}

int main(){
	setdp();
	int n, k;
	while(cin >> n >> k, n+k){
		cout << dp[k][n] << endl;
	}

	return 0;
}
### Prompt

In CPP, your task is to solve the following problem:

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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define rep(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define all(v) (v).begin(), (v).end()
#define rev(s) (s).rbegin(), (s).rend()
#define MP make_pair
#define X first
#define Y second

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int INF = 100000000;

bool isprime[1500];

vector<int> prime;

void setprime(){
	fill(isprime, isprime+1500, true);
	isprime[0] = 	isprime[1] = false;
	for(int i = 2; i < 1500; i++ ){
		if(isprime[i]){
			prime.push_back(i);
			for(int j = i*2; j < 1500; j += i){
				isprime[j] = false;
			}
		}
	}
}

int dp[15][1200] = {};
void setdp(){
	setprime();

	for(int i = 0; i < prime.size(); i++){ // i番目までの素数を使う
		for(int j = 14; j >= 1; j--){	// j個の数字を使う
			if(j == 1) dp[1][prime[i]] += 1;
			for(int k = 0; k+prime[i] < 1200; k++){
				if(!dp[j-1][k]) continue;
				dp[j][k+prime[i]] += dp[j-1][k];
			}
		}
	}
}

int main(){
	setdp();
	int n, k;
	while(cin >> n >> k, n+k){
		cout << dp[k][n] << endl;
	}

	return 0;
}
```