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;
const int maxn = 100010;
long long sum[10000];
long long num[10000];
int main() {
  int s, k;
  scanf("%d%d", &s, &k);
  num[1] = 1;
  sum[1] = 1;
  num[2] = 1;
  sum[2] = 2;
  num[3] = num[1] + num[2];
  sum[3] = 4;
  for (int i = 4; i <= 100; i++) {
    if (i - 1 - k >= 0) {
      num[i] = sum[i - 1] - sum[i - 1 - k];
      sum[i] = sum[i - 1] + num[i];
    } else {
      num[i] = sum[i - 1];
      sum[i] = sum[i - 1] + num[i];
    }
  }
  int ans = 0;
  vector<long long> ret;
  int pos;
  for (int i = 1; i <= 100; i++) {
    if (num[i] > s) {
      pos = i;
      break;
    }
  }
  ret.push_back(0);
  ans = 1;
  for (int i = pos - 1; i >= 1; i--) {
    s -= num[i];
    if (s >= 0) ans++, ret.push_back(num[i]);
    if (s == 0) break;
    if (s < 0) s += num[i];
  }
  printf("%d\n", ans);
  for (int i = 0; i < ret.size(); i++) printf("%I64d ", ret[i]);
  puts("");
  return 0;
}