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;
int modpow(int x, int n) {
  if (n == 0) return 1;
  long long u = modpow(x, n / 2);
  u = (u * u);
  if (n % 2 == 1) u = (u * x);
  return u;
}
const int N = int(1e5) + 10;
const int K = 11;
const int MOD = int(1e9) + 7;
const int INF = int(2e9) + 5;
const long long INF64 = 1e18;
int a[N];
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << "YES" << endl;
    cout << n;
    return;
  }
  if (n < (k * (k + 1)) / 2) {
    cout << "NO" << endl;
    return;
  }
  for (int i = 1; i <= k; i++) {
    a[i] = i;
    n -= i;
  }
  if (n < 0) {
    cout << "NO" << endl;
    return;
  }
  int ex = n / k;
  for (int i = 1; i <= k; i++) a[i] += ex;
  int rem = n % k;
  for (int j = k; j > 0; j--) {
    int t = min(2 * a[j - 1] - a[j], rem);
    a[j] += t;
    rem -= t;
    if (rem == 0) {
      cout << "YES" << endl;
      for (int i = 1; i <= k; i++) cout << a[i] << " ";
      return;
    }
  }
  cout << "NO";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
  return 0;
}
