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
rr = raw_input
rrm = lambda: map(int, raw_input().split())
from collections import deque
def solve(N,K):
    if K == 1: return [N]
    # len(A) == K
    # sum(A) == N
    # A[i] > 0
    # A[i] < A[i+1] <= 2A[i]
    # Lets try to find differences P[i] = A[i+1] - A[i] > 0

    # min sum is 1 + 2 + ... + K
    triangle = K * (K+1) / 2
    if triangle > N: return []
    N -= triangle
    A = range(1, K+1)
    Q, R = divmod(N, K)
    for i in xrange(K):
        A[i] += Q
    i = K-1
    cc = 0
    while R:
        slack = 2 * A[i-1] - A[i]
        delta = min(slack, R)
        R -= delta
        A[i] += delta
        i -= 1
        cc += 1
        if cc == 100000: return []
        if i == 0: i = K-1
        
    return A

ans = solve(*rrm())
if ans:
    print "YES"
    print " ".join(map(str, ans))
else:
    print "NO"
    
