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;
long long ff[111], f[111], ans[111];
int s, k, i, j, ka;
void go(int v, int z) {
  if (z == 0) {
    if (ka == 1) ans[ka++] = 0;
    printf("%d\n", ka);
    for (i = 0; i < ka - 1; i++) printf("%d ", ans[i]);
    printf("%d\n", ans[ka - 1]);
    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() {
  scanf("%d %d", &s, &k);
  f[0] = 1;
  for (i = 1; i <= 50; i++) {
    f[i] = 0;
    for (j = i - 1; j >= i - k && j >= 0; j--) f[i] += f[j];
    if (f[i] > s) break;
  }
  ff[0] = 0;
  for (j = 1; j < i; j++) ff[j] = ff[j - 1] + f[j];
  ka = 0;
  go(i - 1, s);
  return 0;
}