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;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  double n, k;
  cin >> n >> k;
  int fi;
  fi = (n) / k - (1 + k) / 2;
  int left = n - (1 + k + 2 * fi) * k / 2;
  if ((1 + k) * k / 2 > n) {
    cout << "NO";
    return 0;
  } else {
    int accu = 0;
    for (int i = 1; i <= (int)(k); ++i) {
      int h = fi + i + accu;
      if (2 * (h - 1) > h) {
        if (left >= k - i + 1) {
          if (left <= (h - 2) * (k - i + 1)) {
            accu += left / int(k - i + 1);
            left %= left / int(k - i + 1);
          } else {
            accu += h - 2;
            left -= (h - 2) * (k - i + 1);
          }
        }
      }
      ans.push_back(fi + i + accu);
    }
  }
  if (left)
    cout << "NO";
  else {
    cout << "YES" << '\n';
    for (auto e : ans) cout << e << ' ';
  }
}
