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 double pi = acos((long double)-1.0);
const double eps = (double)1e-9;
const int inf = (int)1e9 + 7;
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
  for (auto i : v) os << i << ' ';
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
const int fastio = []() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return 0;
}();
int main(void) {
  int n, k;
  cin >> n >> k;
  if (n < k * 1ll * (k + 1) / 2) {
    return cout << "NO", 0;
  }
  long long int nn = n - k * (k + 1) / 2;
  int a[k];
  for (int i = 0; i < k; i++) {
    a[i] = i + 1 + (nn / k) + (i >= k - (nn % k));
  }
  if (nn != k - 1) {
    cout << "YES" << '\n';
    for_each(a, a + k, [](int a) { cout << a << " "; });
  } else {
    if (k > 3) {
      a[1]--;
      a[k - 1]++;
    }
    if (k == 2 || k == 3) {
      cout << "NO";
      return 0;
    }
    cout << "YES" << '\n';
    for_each(a, a + k, [](int a) { cout << a << " "; });
  }
  return 0;
}
