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;
const int N = 123;
int ka;
long long ff[N], f[N], ans[N];
void go(int v, int z) {
  if (z == 0) {
    if (ka == 1) {
      ans[ka++] = 0;
    }
    cout << ka << '\n';
    for (int i = 0; i <= ka - 1; i++) {
      if (i > 0) cout << ' ';
      cout << ans[i];
    }
    cout << '\n';
    exit(0);
  }
  if (v == 0) {
    return;
  }
  if (ff[v - 1] >= z) {
    go(v - 1, z);
  }
  if (f[v] <= z) {
    ans[ka++] = f[v];
    go(v - 1, z - f[v]);
    ka--;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int s, k;
  cin >> s >> k;
  f[0] = 1;
  int i;
  for (i = 1; i <= 50; i++) {
    for (int j = i - 1; j >= max(i - k, 0); j--) {
      f[i] += f[j];
    }
    if (f[i] > s) {
      break;
    }
  }
  ff[0] = 0;
  for (int j = 1; j < i; j++) {
    ff[j] = ff[j - 1] + f[j];
  }
  ka = 0;
  go(i - 1, s);
  return 0;
}