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 int N = 1e5 + 7;
using namespace std;
long long ans[N];
int main() {
  long long i, j, k, n, m, sum, a0;
  scanf("%lld%lld", &n, &k);
  sum = 0;
  for (i = 1; i <= k; i++) {
    sum = sum * 2 + 1;
    if (sum >= n) break;
  }
  if (sum >= n)
    a0 = 1;
  else
    a0 = n / sum + (n % sum != 0);
  ans[1] = a0;
  m = n;
  m -= a0;
  for (i = 2; i <= k; i++) {
    ans[i] = ans[i - 1] + 1;
    m -= ans[i];
    sum = (ans[i] + 1 + ans[i] + k - i) * (k - i) / 2;
    if (m < sum) {
      ans[i] = n;
      break;
    }
    if (m > sum) {
      long long dx = m - sum;
      if (dx / (k - i + 1) > 2 * ans[i - 1] - ans[i]) {
        m = m - (2 * ans[i - 1] - ans[i]);
        sum = sum + (k - i) * (2 * ans[i - 1] - ans[i]);
        ans[i] = 2 * ans[i - 1];
      } else {
        m = m - dx / (k - i + 1);
        sum = sum + dx / (k - i + 1) * (k - i);
        ans[i] = ans[i] + dx / (k - i + 1);
      }
    } else if (m == sum) {
      for (j = i + 1; j <= k; j++) ans[j] = ans[j - 1] + 1;
      break;
    }
  }
  sum = 0;
  for (i = 1; i <= k; i++) sum += ans[i];
  if (sum != n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (i = 1; i <= k; i++) printf("%lld ", ans[i]);
    printf("\n");
  }
}
