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>
#pragma comment(linker, "/STACK:500000000")
using namespace std;
vector<long long> A, res;
long long s;
int k, size = 1;
int main() {
  scanf("%I64d %d", &s, &k);
  A.push_back(0);
  A.push_back(1);
  while (1) {
    int h = min(k, (int)A.size());
    int it = (int)A.size() - 1;
    long long sum = 0;
    while (h--) {
      sum += A[it--];
      if (sum > 1e9) goto Sol;
    }
    A.push_back(sum);
  }
Sol:
  A.resize(unique(A.begin(), A.end()) - A.begin());
  for (int i = (int)A.size() - 1; i >= 0; i--)
    if (A[i] <= s) {
      res.push_back(A[i]);
      s -= A[i];
    }
  printf("%d\n", res.size());
  printf("%I64d", res[0]);
  for (int i = 1; i < res.size(); i++) printf(" %I64d", res[i]);
  puts("");
  return 0;
}