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 int fib[3000000];
long long int n;
vector<int> v;
int main() {
  int k, i, j;
  scanf("%I64d %d", &n, &k);
  k = min(k, 1000000);
  for (i = 0; i <= k; i++) fib[i] = 0;
  fib[k] = 1;
  for (i = k + 1;; i++) {
    fib[i] = 0;
    for (j = i - 1; j >= i - k; j--) fib[i] += fib[j];
    if (fib[i] > 1000000000) break;
  }
  for (; i >= 0; i--) {
    if (n >= fib[i]) {
      n -= fib[i];
      v.push_back(fib[i]);
    }
    if (n == 0) break;
  }
  printf("%d\n", max((int)v.size(), 2));
  if (v.size() == 1)
    printf("0 %d", v[0]);
  else
    for (i = 0; i < v.size(); i++) printf("%d ", v[i]);
  printf("\n");
  return 0;
}