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;
void a15hx() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long s, k;
  cin >> s >> k;
  if (k < 32) {
    long long arr[1000000];
    for (int i = 1; i <= k - 1; i++) {
      arr[i] = 0;
    }
    arr[k] = 1;
    arr[k + 1] = 1;
    long long curr = 1;
    long long i = k + 2;
    while (curr <= s) {
      arr[i] = curr - arr[i - k - 1] + arr[i - 1];
      curr = arr[i];
      i++;
    }
    vector<long long> ans;
    for (long long j = i - 1; j >= 1; j--) {
      if (arr[j] <= s) {
        ans.push_back(arr[j]);
        s -= arr[j];
        if (arr[j] == 0) break;
      }
    }
    cout << ans.size() << "\n";
    for (auto i : ans) {
      cout << i << " ";
    }
  } else {
    long long arr[32];
    arr[0] = 0;
    for (long long i = 1; i <= 31; i++) {
      arr[i] = pow(2, i - 1);
    }
    vector<long long> ans;
    for (long long j = 31; j >= 0; j--) {
      if (arr[j] <= s) {
        ans.push_back(arr[j]);
        s -= arr[j];
        if (arr[j] == 0) break;
      }
    }
    cout << ans.size() << "\n";
    for (auto i : ans) {
      cout << i << " ";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  a15hx();
  return 0;
}