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, k, na, d[32 + 1];
void nhap() { cin >> n >> k; }
int dem() {
  int i, kq;
  kq = 0;
  for (i = 0; i <= na; i = i + 1) {
    kq = kq + d[i];
  };
  return (kq);
}
int tang() {
  int i;
  for (i = 1; i <= na; i = i + 1) {
    if (d[i] >= 1) {
      d[i] = d[i] - 1;
      d[i - 1] = d[i - 1] + 2;
      break;
    }
  }
}
void xu_ly() {
  int nd;
  nd = n;
  na = -1;
  while (n != 0) {
    na = na + 1;
    int sd;
    sd = n % 2;
    if (sd == 1) {
      d[na] = 1;
    } else {
      d[na] = 0;
    };
    n = n / 2;
  }
  if (dem() > k) {
    cout << "NO";
    return;
  }
  while (dem() < k && d[0] < nd) {
    tang();
  };
  if (dem() < k) {
    cout << "NO";
    return;
  }
  cout << "YES" << endl;
  int i;
  for (i = 0; i <= na; i = i + 1) {
    if (d[i] >= 1) {
      int kq;
      kq = pow(2, i);
      int j;
      for (j = 1; j <= d[i]; j = j + 1) {
        cout << kq << " ";
      };
    }
  }
}
int main() {
  nhap();
  xu_ly();
}