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;
void speed() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(k);
  long long excess = n - ((k * (k + 1)) / 2);
  if (excess < 0) {
    cout << "NO"
         << "\n";
  } else {
    long long add = excess / k;
    excess = excess % k;
    if (k == 1) {
      cout << "YES"
           << "\n";
      cout << n << "\n";
      return;
    }
    if (k <= 3 && excess == k - 1 && add == 0) {
      cout << "NO"
           << "\n";
      return;
    }
    if (k == 2) {
      cout << "YES"
           << "\n";
      if (n % 2 == 0) {
        cout << n / 2 - 1 << " " << n / 2 + 1 << "\n";
      } else {
        cout << n / 2 << " " << n / 2 + 1 << "\n";
      }
      return;
    }
    cout << "YES"
         << "\n";
    for (int i = 0; i < int(k); ++i) {
      a[i] = add + i + 1;
    }
    if (excess == k - 1) {
      a[k - 1]++;
      excess--;
    }
    int j = k - 1;
    while (excess--) {
      a[j]++;
      j--;
    }
    long long sum = 0;
    for (int i = 0; i < int(k); ++i) {
      sum += a[i];
      cout << a[i] << " ";
    }
    cout << "\n";
  }
}
int main() {
  speed();
  int tc = 1;
  while (tc--) {
    solve();
  }
}
