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 gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = powmod(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modif(long long x, long long m) { return (powmod(x, m - 2, m)); }
bool fa(vector<long long> &x, vector<long long> &y) { return x[0] < y[0]; }
bool fa1(vector<long long> &x, vector<long long> &y) { return x[1] < y[1]; }
bool f1(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.second > y.second;
}
bool f2(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.first > y.first;
}
bool f3(long long &x, long long &y) { return x > y; }
const long long maxn = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long b[70];
  for (int i = 0; i < 70; i++) b[i] = 0;
  long long it = 0;
  long long r = 0;
  while (n) {
    long long y = n % 2;
    r += y;
    b[it] = y;
    it++;
    n /= 2;
  }
  if (r > k) {
    cout << "NO\n";
    exit(0);
  }
  r = k - r;
  for (long long i = 69; i >= 1; i--) {
    long long h = min(b[i], r);
    b[i] -= h;
    b[i - 1] += h + h;
    r -= h;
  }
  if (r != 0) {
    cout << "NO\n";
    exit(0);
  }
  cout << "YES\n";
  for (int i = 0; i < 70; i++) {
    for (int t = 0; t < b[i]; t++) {
      long long h = pow(2ll, i);
      cout << h << " ";
    }
  }
  return 0;
}