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>
int n, k;
void otp(int ss, int x);
int main() {
  scanf("%d%d", &n, &k);
  long long sum = 0, sb = 0;
  int i, maxx = 0, j;
  for (i = 1; i <= n; i <<= 1, maxx++)
    if (i & n) {
      sb++;
    }
  if (i > n) {
    maxx--;
  }
  if (sb > k || n < k) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  for (i = maxx; i >= 0; --i) {
    if (n & (1 << i)) {
      sb--;
      if ((1 << i) >= k - sb) {
        otp((1 << i), k - sb);
        break;
      } else {
        k -= (1 << i);
        for (j = 1; j <= (1 << i); j++) printf("1 ");
      }
    }
  }
  for (--i; i >= 0; --i)
    if (n & (1 << i)) {
      printf("%d ", (1 << i));
    }
  return 0;
}
void otp(int ss, int x) {
  int i;
  while (x) {
    if (!(ss % x)) {
      for (i = 1; i <= x; i++) printf("%d ", ss / x);
      break;
    }
    if ((ss >> 1) < x) {
      for (i = 1; i <= (ss >> 1); i++) printf("1 ");
      x -= (ss >> 1);
    } else {
      printf("%d ", ss >> 1);
      x--;
    }
    ss >>= 1;
  }
}