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 a[10005];
int main() {
  long long s;
  int k;
  while (scanf("%I64d%d", &s, &k) != EOF) {
    a[0] = 0;
    a[1] = 1;
    long long x = 0;
    int l = 0, ln;
    for (int i = 2;; i++) {
      a[i] = a[i - 1] + x;
      if (a[i] >= 1000000000) break;
      ln = i;
      x += a[i - 1];
      if (l == k - 1)
        x -= a[i - k];
      else
        l++;
    }
    vector<long long> vec;
    vec.clear();
    for (int i = ln; i > 0; i--) {
      if (a[i] > s) continue;
      s -= a[i];
      vec.push_back(a[i]);
    }
    sort(vec.begin(), vec.end());
    int len = vec.size();
    int d = 0;
    if (len == 1) d = 1;
    printf("%d\n", len + d);
    if (d) printf("0 ");
    for (int i = 0; i < len; i++) printf("%I64d ", vec[i]);
    printf("\n");
  }
  return 0;
}