Problem: A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....

You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).

Output

If it is impossible to represent n as the sum of k powers of two, print NO.

Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.

Examples

Input


9 4


Output


YES
1 2 2 4 


Input


8 1


Output


YES
8 


Input


5 1


Output


NO


Input


3 7


Output


NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[40], d[40];
void insert(int x, int idx) {}
void init() {
  int i;
  a[0] = 1;
  for (i = 1; i <= 30; i++) {
    a[i] = a[i - 1] * 2;
  }
}
void solve() {
  int i;
  if (m > n) {
    printf("NO\n");
    return;
  }
  for (i = 0; i <= 30; i++) {
    d[i] = 0;
  }
  int k = n;
  int cnt = m;
  for (i = 30; i >= 0; i--) {
    if (k >= a[i]) {
      k -= a[i];
      d[i] = 1;
      cnt--;
    }
  }
  if (cnt < 0) {
    printf("NO\n");
    return;
  }
  for (i = 30; i > 0 && cnt > 0; i--) {
    if (d[i] == 0) continue;
    if (cnt >= a[i] - 1) {
      cnt -= (a[i] - 1);
      d[i]--;
      d[0] += (a[i]);
    }
    while (d[i] && cnt > 0) {
      cnt--;
      d[i]--;
      d[i - 1] += 2;
    }
  }
  printf("YES\n");
  for (i = 0; i <= 30; i++) {
    if (d[i] == 0) continue;
    while (d[i]) {
      d[i]--;
      printf("%d ", a[i]);
    }
  }
  printf("\n");
}
int main() {
  init();
  while (scanf("%d %d", &n, &m) != EOF) {
    solve();
  }
  return 0;
}