Well-known Numbers

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 a ₁, a ₂, …, a _m. 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:

#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <sstream>

using namespace std;

long long ff[111], f[111], ans[111];
int s, k, i, j, ka;

void go(int v, int z) {
if (z == 0) {
if (ka == 1) ans[ka++] = 0;
printf("%d\n",ka);
for (i=0;i<ka-1;i++) printf("%d ",ans[i]);
    printf("%d\n",ans[ka-1]);
    exit(0);
  }
  if (v == 0) return;
  if (ff[v-1] >= z) go(v-1,z);
if (f[v] <= z) {
    ans[ka++] = f[v];
    go(v-1,z-f[v]);
    ka--;
  }
}

int main() {
//  freopen("in","r",stdin);
//  freopen("out","w",stdout);
  scanf("%d %d",&s,&k);
  f[0] = 1;
  for (i=1;i<=50;i++) {
    f[i] = 0;
    for (j=i-1;j>=i-k && j>=0;j--) f[i] += f[j];
if (f[i] > s) break;
}
ff[0] = 0;
for (j=1;j<i;j++) ff[j] = ff[j-1]+f[j];
  ka = 0;
  go(i-1,s);
  return 0;
}