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;
void islam() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
}
map<long long, long long> dp, mp;
void sm(long long i, long long j) {
  long long x = i;
  while (i > j) {
    i--;
    dp[x] += dp[i];
  }
}
vector<long long> V, v;
bool ok = 0;
void bc(long long x, long long ans) {
  if (ok) return;
  if (ans < 0) return;
  if (!ans) {
    ok = 1;
    if (V.size() == 1) {
      cout << "2\n0 " << V[0];
      return;
    }
    cout << V.size() << endl;
    cout << V[0];
    for (int i = 1; i < V.size(); i++) cout << " " << V[i];
    return;
  }
  for (int i = x; i < v.size(); i++) {
    V.push_back(v[i]);
    bc(i + 1, ans - v[i]);
    V.pop_back();
  }
}
int main() {
  islam();
  long long s, k, a = 1;
  cin >> s >> k;
  dp[k] = 1;
  for (int i = k + 1;; i++) {
    if (i > k * 2) a -= dp[i - k - 1];
    dp[i] = a;
    a += dp[i];
    if (!mp[dp[i]]) v.push_back(dp[i]);
    mp[dp[i]]++;
    if (dp[i] == s) return cout << "2\n0 " << s << endl, 0;
    if (dp[i] > s || a > s) break;
  }
  sort(v.rbegin(), v.rend());
  bc(0, s);
  return 0;
}