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;
mt19937 rng((uint64_t)chrono::duration_cast<chrono::nanoseconds>(
                chrono::high_resolution_clock::now().time_since_epoch())
                .count());
inline int rand(int l, int r) {
  uniform_int_distribution<int> RNG(l, r);
  return RNG(rng);
}
const int N = 1e5 + 5;
int64_t x[105] = {0};
vector<int> arr{0, 1};
vector<int> ans;
void f(int i, int sum) {
  if (sum < 0 || i >= (int)arr.size()) return;
  if (sum == 0) {
    printf("%d\n", (int)ans.size());
    for (int j : ans) printf("%d ", j);
    printf("\n");
    exit(0);
  }
  ans.push_back(arr[i]);
  f(i + 1, sum - arr[i]);
  ans.pop_back();
  f(i + 1, sum);
}
void solve(int casenum) {
  int n, k;
  scanf("%d%d", &n, &k);
  x[1] = 1;
  for (int i = 2;; i++) {
    int64_t add = 0;
    for (int j = 0; j < min(i - 1, k); j++) add += x[i - j - 1];
    x[i] = add;
    arr.push_back(add);
    if (x[i] >= n) break;
  }
  sort(arr.begin(), arr.end());
  arr.erase(unique(arr.begin(), arr.end()), arr.end());
  reverse(arr.begin(), arr.end());
  ans.push_back(0);
  f(0, n);
}
int main() {
  int T = 1;
  for (int i = 1; i <= T; i++) solve(i);
  return 0;
}
