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 MAX_N = 1000;
int fs[MAX_N], as[MAX_N];
int main() {
  int s, k;
  scanf("%d%d", &s, &k);
  int n = 2;
  fs[0] = 0, fs[1] = 1;
  while (n < MAX_N) {
    fs[n] = 0;
    for (int i = max(0, n - k); i < n; i++) fs[n] += fs[i];
    if (fs[n] > s) break;
    n++;
  }
  int m = 0;
  while (s > 0) {
    while (n > 0 && s < fs[n - 1]) n--;
    s -= fs[--n];
    as[m++] = fs[n];
  }
  if (m == 1) as[m++] = 0;
  printf("%d\n", m);
  for (int i = m - 1; i >= 0; i--) {
    printf("%d", as[i]);
    putchar(i ? ' ' : '\n');
  }
  return 0;
}