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 long long maxn = 1e6 + 9;
long long n, k;
long long a[maxn];
int main() {
  cin >> n >> k;
  long long ret(0);
  for (long long i = 1; i <= k; ++i) a[i] = i, ret += a[i];
  if (ret > n) {
    puts("NO");
    return 0;
  }
  long long tmp(n - ret);
  if (tmp >= k) {
    long long lun(tmp / k);
    for (long long i = 1; i <= k; ++i) a[i] += lun;
    ret += lun * k;
  }
  long long up(0);
  while (true) {
    tmp = n - ret;
    if (!tmp) break;
    long long now(std::max(k - tmp + 1, up));
    if (a[now - 1] * 2 < a[now] + 1) {
      long long f(0);
      for (long long j = now + 1; j <= k; ++j) {
        if (a[j - 1] * 2 >= a[j] + 1) {
          f = j;
          break;
        }
      }
      if (!f) {
        puts("NO");
        return 0;
      }
      up = f;
      for (long long j = f; j <= k; ++j) ++a[j];
      ret += k - f + 1;
    } else {
      for (long long i = now; i <= k; ++i) ++a[i];
      ret += k - now + 1;
    }
  }
  puts("YES");
  for (long long i = 1; i <= k; ++i) cout << a[i] << ' ';
  return 0;
}
