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>
const double eps = 0.0000001;
using namespace std;
inline int sgn(double x) { return (x > eps) - (x < -eps); }
inline char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
inline int read() {
  register int x = 0, f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x * f;
}
long long f[(long long)1e5];
long long num[(long long)1e5];
long long Idx;
bool flag = 1;
void sum(long long Sum, long long idx, long long IDX) {
  if (Sum < 0) return;
  if (idx <= 0) return;
  if (Sum == 0) {
    Idx = IDX;
    flag = 0;
    return;
  }
  if (flag == 0) return;
  num[IDX] = f[idx];
  sum(Sum - f[idx], idx - 1, IDX + 1);
  sum(Sum, idx - 1, IDX);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n = 3;
  f[1] = 1;
  f[2] = 1;
  long long s, k;
  cin >> s >> k;
  for (; f[n - 1] < s; ++n) {
    f[n] = 2 * f[n - 1] - (n - k - 1 < 0 ? 0 : f[n - k - 1]);
  }
  sum(s, n, 1);
  cout << Idx - 1 << endl;
  if (Idx - 1 == 1) {
    cout << 0 << " ";
  }
  for (int i = 1; i < Idx; ++i) cout << num[i] << " ";
}