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>
using namespace std;
long long m, n, a[100010], s;
bool ok = true;
int main() {
  cin.tie(0), cout.sync_with_stdio(false);
  cin >> m >> n;
  if (n == 1) {
    cout << "YES\n" << m << '\n';
    return 0;
  }
  s = m - ((n + 1) * n / 2);
  if (s < 0) ok = false;
  for (int i = 1; i <= n; i++) {
    a[i] = s / n;
    a[i] += i;
    if (i > n - s % n) a[i] += 1;
  }
  if (s == n - 1) {
    if (n == 2 || n == 3) ok = false;
    a[2] -= 1;
    a[n] += 1;
  }
  if (!ok) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
  cout << '\n';
  return 0;
}
