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 a, b;
priority_queue<long long> pq;
void shitweight() {
  long long aa = a, bb = 1;
  int i = 0;
  while (aa) {
    if (aa & 1) pq.push(bb);
    aa >>= 1;
    bb <<= 1;
  }
}
void printshit() {
  long long aa = a;
  long long shit = 1;
  int fi = 1;
  while (aa) {
    if (aa & 1) {
      if (fi)
        fi = 0;
      else
        printf(" ");
      cout << shit;
      shit <<= 1;
    }
    aa >>= 1;
  }
}
int main() {
  while (cin >> a >> b) {
    while (!pq.empty()) pq.pop();
    shitweight();
    while (pq.size() < b) {
      if (pq.top() == 1) break;
      long long k = (pq.top() >> 1);
      pq.pop();
      pq.push(k);
      pq.push(k);
    }
    if (pq.size() > b || pq.size() < b) {
      cout << "NO" << endl;
      continue;
    } else if (pq.size() == b) {
      cout << "YES" << endl;
      cout << pq.top();
      pq.pop();
      while (!pq.empty()) {
        cout << " " << pq.top();
        pq.pop();
      }
      cout << endl;
      continue;
    }
  }
}