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
#include <bits/stdc++.h>
using namespace std;
long long sum = 0;
long long abn[101];
bool used[101];
long long sd[101];
bool dnf(vector<long long> &ivec, long long i, long long k, long long s,
         long long &sum) {
  if (s == 0) return true;
  for (long long j = i; j >= 0; j--) {
    if (used[j] == true) continue;
    if (ivec[j] > s) continue;
    used[j] = true;
    sd[sum++] = j + 1;
    if (dnf(ivec, i, k, s - ivec[j], sum)) return true;
    used[j] = false;
    sd[j] = 0;
  }
  return false;
}
int main() {
  long long s, k;
  while (cin >> s >> k) {
    long long i;
    if (s == 1) {
      cout << 2 << endl;
      cout << 0 << " " << 1 << endl;
      continue;
    }
    memset(abn, 0, sizeof(abn));
    abn[0] = 1;
    sum = 0;
    memset(used, 0, sizeof(used));
    memset(sd, 0, sizeof(sd));
    vector<long long> ivec;
    ivec.push_back(1);
    long long er = 1;
    for (i = 1; abn[i - 1] < s; i++) {
      if (abn[i] == 0) {
        for (long long e = i - 1; e >= i - k && e >= 0; e--) {
          abn[i] += abn[e];
        }
      }
      if (count(ivec.begin(), ivec.end(), abn[i]) == 0) {
        ivec.push_back(abn[i]);
        er++;
      }
    }
    if (dnf(ivec, er - 1, k, s, sum)) {
      cout << sum + 1 << endl;
      cout << 0;
      for (long long d = 0; d != sum; d++) {
        cout << " " << ivec[sd[d] - 1];
      }
    }
    cout << endl;
  }
  return 0;
}
