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 main() {
  int N, K, len, counter = 0;
  long long kbonacci[110], answer[110];
  memset(kbonacci, 0, sizeof(kbonacci));
  scanf("%d %d", &N, &K);
  kbonacci[0] = 1;
  for (int i = 1; kbonacci[i - 1] < N; i++) {
    for (int j = i - 1; (j >= 0 && j >= (i - K)); j--) {
      kbonacci[i] += kbonacci[j];
    }
    len = i;
  }
  for (int i = len; i >= 0; i--) {
    if (kbonacci[i] <= N) {
      answer[counter++] = kbonacci[i];
      N -= kbonacci[i];
    }
  }
  printf("%d\n", counter + 1);
  for (int i = 0; i < counter; i++) {
    printf("%d ", answer[i]);
  }
  printf("0\n");
  return 0;
}