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 pw(long long bs, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans *= bs;
    bs *= bs;
    x >>= 1;
  }
  return ans;
}
long long res[100005], x, n;
int main() {
  cin >> x >> n;
  if (n == 1) {
    puts("YES");
    printf("%lld", x);
    return 0;
  }
  for (long long a = 1;; a++) {
    if (n * a + n * (n - 1) / 2 > x) {
      a--;
      if (a <= 0) break;
      if (n <= 30 && a * (pw(2, n) - 1) < x) break;
      puts("YES");
      res[1] = a;
      for (int i = 2; i <= n; i++) res[i] = res[i - 1] + 1;
      long long nd = x - (n * a + n * (n - 1) / 2);
      while (nd > 0) {
        for (int i = n; i >= 2; i--) {
          long long chg = min(nd, res[i - 1] * 2 - res[i]);
          res[i] += chg;
          nd -= chg;
        }
      }
      for (int i = 1; i <= n; i++) printf("%lld ", res[i]);
      puts("");
      return 0;
    }
  }
  puts("NO");
}
