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 double pi = 2 * acos(0.0);
const int maxn = 2e5 + 10;
vector<int> ans;
int main() {
  int n, k;
  cin >> n >> k;
  int l = 1, r = 1e9;
  while (l <= r) {
    int mid = (l + r) >> 1;
    long long res = 1LL * (mid + mid + k - 1) * k / 2;
    if (res <= n)
      l = mid + 1;
    else
      r = mid - 1;
  }
  long long res = 1LL * (r + r + k - 1) * k / 2;
  long long sub = n - res;
  if (sub < 0) {
    printf("NO\n");
    return 0;
  } else if (sub == 0) {
    for (int i = 1; i <= k; i++) ans.push_back((int)(i + r - 1));
  } else if (sub == k - 1) {
    if (k == 2)
      ans.push_back(r), ans.push_back((int)(r + 2));
    else {
      long long sum = 0;
      for (int i = 1; i <= k - 2; i++) {
        ans.push_back((int)(i + r - 1));
        sum += i + r - 1;
      }
      ans.push_back(k + r - 1);
      sum += r + k - 1;
      ans.push_back((int)(n - sum));
    }
  } else {
    long long sum = 0;
    for (int i = 1; i <= k - 1; i++) {
      ans.push_back((int)(i + r - 1));
      sum += i + r - 1;
    }
    ans.push_back(n - sum);
  }
  int flag = 1;
  for (int i = 0; i < ans.size(); i++) {
    if (i != 0 && (ans[i] > 2 * ans[i - 1] || ans[i] <= ans[i - 1])) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    printf("YES\n");
    for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  } else
    printf("NO\n");
  return 0;
}
