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 dp[1000005];
int main() {
  dp[1] = 0;
  dp[2] = 1;
  long long s, k;
  cin >> s >> k;
  for (int i = 3; i <= 100; i++) {
    for (int j = i - 1; j >= max(i - k, (long long)1); j--) {
      dp[i] += dp[j];
    }
  }
  vector<long long> v;
  for (int i = 55; i >= 1; i--) {
    if (dp[i] <= s) {
      s -= dp[i];
      v.push_back(dp[i]);
    }
  }
  sort(v.begin(), v.end());
  cout << v.size() << endl;
  for (auto i : v) {
    cout << i << " ";
  }
}