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;
int s, k, i, j;
int a[100];
int main() {
  while (~scanf("%d%d", &s, &k)) {
    memset(a, 0, sizeof(a));
    a[0] = 0;
    a[1] = 1;
    for (i = 2;; i++) {
      for (j = 1; j <= k && j <= i; j++) a[i] += a[i - j];
      if (a[i] > s) break;
    }
    int t = i, ans = 0, n = s;
    for (i = t; i > 0; i--) {
      if (n >= a[i]) {
        n -= a[i];
        ans++;
      }
    }
    printf("%d\n", ans + 1);
    cout << 0;
    for (i = t; i > 0; i--)
      if (s >= a[i]) {
        s -= a[i];
        printf(" %d", a[i]);
      }
    cout << endl;
  }
  return 0;
}