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 int calculation(long long int n) {
  if (n < 0) return 0;
  return (n * (n + 1)) / 2;
}
long long int A[100005];
int main() {
  long long int k, n, tut = 0;
  cin >> k >> n;
  long long int b = 0, e = 10000000000;
  while (b <= e) {
    long long int mid = (b + e) / 2;
    if (calculation(mid) - calculation(mid - n) <= k) {
      tut = max(tut, mid);
      b = mid + 1;
    } else
      e = mid - 1;
  }
  long long int temp = k - calculation(tut) + calculation(tut - n);
  for (int i = n; i >= 1; i--) {
    A[i] = tut;
    tut--;
  }
  for (int i = n; i >= 2; i--) {
    A[i] += min(A[i - 1] - 1, temp);
    temp -= min(A[i - 1] - 1, temp);
  }
  A[1] += temp;
  bool ans = true;
  for (int i = 2; i <= n; i++) {
    if (A[i] <= A[i - 1] || A[i] > 2 * A[i - 1]) ans = false;
  }
  if (!ans) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << A[i] << " ";
    cout << endl;
  }
  return 0;
}
