Polycarp has to solve exactly n problems to improve his programming skill before an important programming competition. But this competition will be held very soon, most precisely, it will start in k days. It means that Polycarp has exactly k days for training!

Polycarp doesn't want to procrastinate, so he wants to solve at least one problem during each of k days. He also doesn't want to overwork, so if he solves x problems during some day, he should solve no more than 2x problems during the next day. And, at last, he wants to improve his skill, so if he solves x problems during some day, he should solve at least x+1 problem during the next day.

More formally: let [a_1, a_2, ..., a_k] be the array of numbers of problems solved by Polycarp. The i-th element of this array is the number of problems Polycarp solves during the i-th day of his training. Then the following conditions must be satisfied: 

  * sum of all a_i for i from 1 to k should be n; 
  * a_i should be greater than zero for each i from 1 to k; 
  * the condition a_i < a_{i + 1} ≤ 2 a_i should be satisfied for each i from 1 to k-1. 



Your problem is to find any array a of length k satisfying the conditions above or say that it is impossible to do it.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10^5) — the number of problems Polycarp wants to solve and the number of days Polycarp wants to train.

Output

If it is impossible to find any array a of length k satisfying Polycarp's rules of training, print "NO" in the first line.

Otherwise print "YES" in the first line, then print k integers a_1, a_2, ..., a_k in the second line, where a_i should be the number of problems Polycarp should solve during the i-th day. If there are multiple answers, you can print any.

Examples

Input


26 6


Output


YES
1 2 4 5 6 8 


Input


8 3


Output


NO


Input


1 1


Output


YES
1 


Input


9 4


Output


NO
rint = lambda: int(input())
rmint = lambda: map(int, input().split())
rlist = lambda: list(rmint())

n, k = rmint()
t = k
a = [0] * k
def no(): print("NO"); exit(0)
d = [-1] * (k+1); d[0] = 1
mx = 10 ** 9
for i in range(1, k+1):
    if d[i-1] * 2 > mx: break
    d[i] = d[i-1] * 2
for i in range(k):
    if n < 0: break
    if d[t] < 0:
        p = 1
        if i: p = a[i-1] + 1
    else:
        f = d[t] - 1
        p = max(1, (n + f - 1) // f)
    if i:
        p = max(p, a[i-1] + 1)
        p = min(p, a[i-1] * 2)
    a[i] = p
    t -= 1
    n -= p
if n: no()
print("YES")
for i in range(k): print(a[i],end=' ')
