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 int decToBinary(int n) {
  int binaryNum[32];
  long long int cnt = 0;
  int i = 0;
  while (n > 0) {
    binaryNum[i] = n % 2;
    if (n % 2 == 1) cnt++;
    n = n / 2;
    i++;
  }
  long long int k = 0;
  long long int sum = 0;
  for (int j = i - 1; j >= 0; j--) {
    if (binaryNum[j]) {
      sum += pow(2, k);
    }
    k++;
  }
  return cnt;
}
void solve() {
  long long int a, b, c, d, m, n, p, q, r, x, y, z, aa, bb, cc, dd;
  long long int i, j, k, l;
  long long int cnt = 0, cnt1 = 0, cnt2 = 0, cnt3 = 0, sum = 0, sum2 = 0,
                mx = LLONG_MIN, mn = LLONG_MAX;
  long long int flag = 0, flag2 = 0;
  vector<long long int> vec1, vec2, vec3;
  vector<long long int>::iterator it, it2;
  string s1, s2, s3;
  cin >> n >> k;
  long long int ff = n;
  z = 10;
  j = 1;
  for (long long int i = (long long int)(0); i < (long long int)(30); ++i) {
    vec1.push_back(j);
    j *= 2;
  }
  mn = decToBinary(n);
  if (k < mn || k > ff)
    cout << "NO\n";
  else {
    cout << "YES\n";
    while (n != k) {
      it = upper_bound(vec1.begin(), vec1.end(), n);
      for (it2 = it; it >= vec1.begin(); it2--) {
        x = n - *it2;
        if (x >= k - 1) {
          cout << *it2 << " ";
          n -= *it2;
          k--;
          break;
        }
      }
    }
    for (long long int i = (long long int)(0); i < (long long int)(n); ++i)
      cout << "1 ";
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}