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>
const long double eps = 1e-9;
const long long mod = 1e9 + 7, ll_max = (long long)1e18;
const int MX = 2e5 + 10, int_max = 0x3f3f3f3f;
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long arr[MX];
int solve() {
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    arr[i] += i + 1;
    n -= i + 1;
  }
  if (n < 0) return 0;
  for (int i = 0; i < k; i++) {
    arr[i] += (n / k);
  }
  n %= k;
  for (int i = k - 1; i > 0; i--) {
    int del = min(2 * arr[i - 1] - arr[i], n);
    n -= del;
    arr[i] += del;
  }
  return (n == 0);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  if (solve()) {
    cout << "YES\n";
    for (int i = 0; i < k; i++) cout << arr[i] << " ";
    cout << "\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}
