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>
#pragma warning(disable : 4996)
using namespace std;
int n, k;
int cnt, num;
struct node {
  bool operator()(int x, int y) { return x > y; }
};
priority_queue<int, vector<int>, node> que2;
queue<int> que;
int main() {
  scanf("%d%d", &n, &k);
  int tmp = n;
  while (tmp) {
    if (tmp & 1) {
      que.push(pow(2.0, num));
      cnt++;
    }
    num++;
    tmp >>= 1;
  }
  if (k > n || k < cnt)
    printf("NO\n");
  else {
    int count = cnt;
    while (count < k) {
      int f = que.front();
      que.pop();
      if (f > 1) {
        que.push(f / 2);
        que.push(f / 2);
        count++;
      } else {
        que.push(f);
      }
    }
    while (!que.empty()) {
      int ans = que.front();
      que.pop();
      que2.push(ans);
    }
    printf("YES\n");
    while (!que2.empty()) {
      int ans = que2.top();
      que2.pop();
      printf("%d ", ans);
    }
    printf("\n");
  }
  return 0;
}