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
#include <bits/stdc++.h>
const int N = 1e6 + 10;
const int mod = 998244353;
using namespace std;
int a[N];
int cal(int n, int k) {
  long long t = n - 1LL * (k - 1) * k / 2, u = t / k;
  return u;
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long d, u;
  if (k > 30)
    d = 1;
  else {
    long long t = (1 << k) - 1;
    if (n % t)
      d = n / t + 1;
    else
      d = n / t;
  }
  long long t = n - 1LL * (k - 1) * k / 2;
  u = t / k;
  if (d > u) return cout << "NO" << endl, 0;
  cout << "YES" << endl;
  int s = n;
  a[1] = u, s -= u;
  for (int i = 2; i < k; ++i)
    a[i] = min(2 * a[i - 1], cal(s, k - i + 1)), s -= a[i];
  if (k > 1) a[k] = s;
  for (int i = 1; i <= k; ++i) cout << a[i] << " ";
  return 0;
}
