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 oo = 1e9;
const long long OO = 1ll << 62;
map<long long, bool> use;
int s, k;
vector<long long> K, res;
int main() {
  scanf("%d%d", &s, &k);
  K.resize(51);
  K[0] = 0;
  K[1] = 1;
  for (int i = 2; i <= 50; ++i) {
    K[i] = 0;
    for (int j = i - 1; j >= i - k && j >= 0; --j) K[i] += K[j];
  }
  K.resize(unique(K.begin(), K.end()) - K.begin());
  res.push_back(0);
  for (int i = 50; i > 0; --i) {
    if (K[i] > s) continue;
    s -= K[i];
    res.push_back(K[i]);
  }
  printf("%d\n", (int)res.size());
  for (int i = 0; i < (int)res.size(); ++i) printf("%lld ", res[i]);
  return 0;
}