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
def first_fib(s,k):
    bon = [0,1]
    summ = 0
    while summ < s:
        summ = 0
        if len(bon) < k: count = len(bon)
        else: count = k
        last = bon.index(max(bon))
        summ = sum(bon[-count:-1])+bon[last]
        bon.append(summ)
    del bon[last+1]
    return bon
bon = [0]
i = 0
s,k = map(int,raw_input().split())
if s == 1: print "2" + "\n" + "1 0"
elif s == 2: print "2" + "\n" + "2 0"
elif s == 3: print "2" + "\n" + "1 2"
elif s == 4: print "2" + "\n" + "4 0"
else:
    if k<32: bon = first_fib(s,k)
    else:
        while (2**i) < s:
            bon.append(2**i)
            i+=1
    arr = []
    while s != 0:
        if s >= bon[-1]:
            s-=bon[-1]
            arr.append(bon.pop())
        else:
            del bon[-1]
    l = len(arr)
    print l
    print ' '.join(map(str,arr))
