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 dp[200];
int main() {
  int i, j, k, n, s;
  dp[0] = dp[1] = 1;
  cin >> s >> k;
  for (i = 2;; i++) {
    for (j = max(i - k, 0); j < i; j++) dp[i] += dp[j];
    if (dp[i] > s) break;
  }
  i--;
  vector<int> an;
  while (s) {
    if (s >= dp[i]) {
      an.push_back(dp[i]);
      s -= dp[i];
    }
    i--;
  }
  an.push_back(0);
  printf("%d\n", an.size());
  for (i = 0; i < an.size(); i++) {
    if (i) printf(" ");
    printf("%d", an[i]);
  }
  puts("");
  return 0;
}