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;
long long k, s, sz;
vector<long long> f;
vector<long long> sum;
vector<long long> ans;
int main() {
  scanf("%I64d%I64d", &s, &k);
  f.push_back(1LL);
  sum.push_back(1LL);
  for (long long i = 1; i; i++) {
    long long x = sum[i - 1], y = 0;
    if (i - k - 1 >= 0) y = sum[i - k - 1];
    if (x - y > s) break;
    f.push_back(x - y);
    sum.push_back(sum[i - 1] + x - y);
    sz++;
  }
  while (s) {
    long long l = 0, r = sz, last;
    while (l <= r) {
      long long m = (l + r) >> 1;
      if (f[m] <= s) {
        last = m;
        l = m + 1;
      } else
        r = m - 1;
    }
    ans.push_back(f[last]);
    s -= f[last];
  }
  long long ans2 = ans.size();
  if (ans2 == 1) {
    printf("2\n");
    printf("0 %I64d\n", f[sz]);
  } else {
    long long ans_sz = ans.size();
    printf("%I64d\n", ans_sz);
    for (long long i = 0; i < ans_sz; i++) printf("%I64d ", ans[i]);
    printf("\n");
  }
  return 0;
}