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;
long long n, t, a, sum, b, ar[1005], c, lala, o, g, br[1005];
void cek(long long x, long long &y, long long z, long long c) {
  if (c < 0) {
    return;
  }
  if (y == z) {
    lala = 0;
    return;
  } else if (y < z && c != 0) {
    ar[c]--;
    ar[c - 1] += 2;
    y++;
    cek(x, y, z, c - 1);
    cek(x, y, z, c - 1);
  }
}
int main() {
  scanf("%lld %lld", &t, &n);
  a = t;
  sum = 0;
  b = 0;
  while (a > 0) {
    sum = sum + a % 2;
    ar[b] = a % 2;
    b++;
    a = a / 2;
  }
  b--;
  lala = 1;
  for (int i = b; i >= 0; i--) {
    if (ar[i] == 1) {
      cek(t, sum, n, i);
    }
  }
  if (lala == 0) {
    printf("YES\n");
    for (int i = 0; i <= b; i++) {
      if (ar[i] > 0) {
        for (int j = 1; j <= ar[i]; j++) {
          g = 1;
          for (int k = 0; k < i; k++) {
            g = g * 2;
          }
          printf("%lld ", g);
        }
      }
    }
    printf("\n");
  } else if (lala == 1) {
    printf("NO\n");
  }
  return 0;
}