Problem: Numbers k-bonacci (k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:

  * F(k, n) = 0, for integer n, 1 ≤ n < k; 
  * F(k, k) = 1; 
  * F(k, n) = F(k, n - 1) + F(k, n - 2) + ... + F(k, n - k), for integer n, n > k. 



Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.

You've got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers. 

Input

The first line contains two integers s and k (1 ≤ s, k ≤ 109; k > 1).

Output

In the first line print an integer m (m ≥ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a1, a2, ..., am. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.

It is guaranteed that the answer exists. If there are several possible answers, print any of them.

Examples

Input

5 2


Output

3
0 2 3


Input

21 5


Output

3
4 1 16

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
map<long long, long long> arr;
int pre[60] = {1};
vector<int> vec;
int main() {
  int s, k;
  scanf("%d %d", &s, &k);
  long long sum = 1;
  arr[k] = 1;
  int idx = k + 1;
  for (; 1; idx++) {
    sum -= arr[idx - k - 1];
    arr[idx] = sum;
    sum += arr[idx];
    if (sum > 1e9 + 7) break;
  }
  vec.push_back(0);
  for (int i = idx; i; i--) {
    if (s >= arr[i]) {
      vec.push_back(arr[i]);
      s -= arr[i];
    }
    if (s == 0) break;
  }
  printf("%d\n", vec.size());
  for (int i = 0; i < vec.size(); i++) printf("%d ", vec[i]);
}