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
#include <bits/stdc++.h>
using namespace std;
const int P = 4;
const int MAXP = 470;
const int MAXAN = 60;
int ps[] = {2, 3, 7, 11, 13, 17, 229, 281, 349, 409, 463};
long long a[MAXAN];
int na;
int g[MAXAN][P][MAXP];
long long s, k;
void get_a() {
  int i;
  a[0] = 1;
  a[1] = 1;
  for (i = 1; a[i] <= s; i++) {
    a[i + 1] = a[i] * 2;
    if (i >= k) a[i + 1] -= a[i - k];
  }
  na = i;
  for (i = 0; i < na - 1; i++) a[i] = a[i + 1];
  na--;
}
bool can_make(int i, long long s1) {
  int j;
  for (j = 0; j < P; j++)
    if (not g[i + 1][j][s1 % ps[j]]) return false;
  return true;
}
void solve() {
  if (na == 0) {
    cout << "2\n1 0\n";
    return;
  }
  int i, j, x;
  i = 0;
  for (j = 0; j < P; j++)
    for (x = 0; x < ps[j]; x++) g[i][j][x] = 0;
  for (j = 0; j < P; j++) g[i][j][0] = 2;
  for (i = 0; i < na; i++) {
    for (j = 0; j < P; j++)
      for (x = 0; x < ps[j]; x++) {
        g[i + 1][j][x] = 0;
        if (g[i][j][(x + ps[j] - a[i]) % ps[j]]) g[i + 1][j][x] |= 1;
        if (g[i][j][x]) g[i + 1][j][x] |= 2;
      }
  }
  vector<int> solu;
  long long cur = s;
  long long s1;
  i = na - 1;
  while (cur > 0) {
    s1 = cur - a[i];
    if (can_make(i - 1, s1)) {
      solu.push_back(a[i]);
      cur -= a[i];
    } else {
    }
    i--;
  }
  if (solu.size() == 1) solu.push_back(0);
  cout << solu.size() << endl;
  for (i = 0; i < solu.size(); i++)
    cout << solu[i] << (i == solu.size() - 1 ? '\n' : ' ');
}
int main() {
  cin >> s >> k;
  get_a();
  solve();
  return 0;
}
