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 s, k, p;
unsigned long long fib[100];
int answer[100];
void solve(int pos, int k, unsigned long long sum) {
  if (sum > s) {
    return;
  }
  if (sum == s) {
    cout << k + 1 << endl;
    for (int i = 1; i <= k; i++) {
      cout << answer[i] << ' ';
    }
    cout << 0;
    exit(0);
  }
  {
    answer[k + 1] = fib[pos];
    if (pos) solve(pos - 1, k + 1, sum + fib[pos]);
    if (pos) solve(pos - 1, k, sum);
  }
}
int main() {
  cin >> s >> k;
  fib[1] = 1;
  for (int i = 2;; i++) {
    for (int j = i - 1; j >= max(0, i - k); j--) {
      fib[i] += fib[j];
    }
    if (fib[i] >= 1000000000) {
      p = i;
      break;
    }
  }
  solve(p, 0, 0);
  return 0;
}