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>
#pragma comment(linker, "/STACK:16777216")
const int inf = 2147383647;
const double pi = 2 * acos(0.0);
const double eps = 1e-7;
const int maxint = 2147483647;
const int minint = -2147483648;
using namespace std;
int MIN(int a, int b) {
  if (a < b) return a;
  return b;
}
int MAX(int a, int b) {
  if (a > b) return a;
  return b;
}
vector<int> grid[35];
void init() {
  int i, j, k;
  long long num;
  for (i = 2; i <= 32; i++) {
    grid[i].push_back(0);
    for (j = 1;; j++) {
      if (j < i)
        grid[i].push_back(0);
      else if (j == i)
        grid[i].push_back(1);
      else {
        num = 0;
        for (k = 1; k <= i; k++) {
          num += grid[i][j - k];
        }
        if (num > 1000000000)
          break;
        else
          grid[i].push_back(num);
      }
    }
  }
}
int main() {
  init();
  int sum, k;
  while (scanf("%d %d", &sum, &k) != EOF) {
    int num = 0;
    int arr[50];
    int i = 0, temp = sum;
    if (k > 32) {
      i = 0;
      while (temp) {
        if (temp % 2 == 0)
          arr[i++] = 0;
        else {
          arr[i++] = 1;
          num++;
        }
        temp /= 2;
      }
      if (num > 1) {
        printf("%d\n", num);
        int flag = 0;
        for (i = 0; i < 40; i++) {
          if (arr[i] == 1) {
            if (flag == 1) printf(" ");
            printf("%d", (int)pow(2, i));
            flag = 1;
          }
        }
        printf("\n");
        continue;
      } else {
        printf("%d\n", num + 1);
        printf("0");
        for (i = 0; i < 40; i++) {
          if (arr[i] == 1) {
            printf(" %d", (int)pow(2, i));
          }
        }
        printf("\n");
        continue;
      }
    }
    stack<int> s;
    for (i = grid[k].size() - 1; i >= 0; i--) {
      if (grid[k][i] <= temp) {
        num++;
        temp -= grid[k][i];
        s.push(grid[k][i]);
      }
      if (temp == 0) break;
    }
    if (num > 1) {
      printf("%d\n", num);
      int flag = 0;
      while (!s.empty()) {
        temp = s.top();
        s.pop();
        if (flag == 1) printf(" ");
        printf("%d", temp);
        flag = 1;
      }
    } else {
      printf("%d\n", num + 1);
      printf("0");
      while (!s.empty()) {
        temp = s.top();
        s.pop();
        printf(" %d", temp);
      }
    }
  }
  return 0;
}