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;
const int MAX = 1e6 + 10;
const int OO = 0x3f3f3f3f;
const double EPS = 1e-6;
const long long MOD = 1000000000;
long long n, k, ans, p;
long long cnt[MAX];
int32_t main() {
  cin >> n >> k;
  for (long long i = 32; i >= 0; i--)
    if (n & (1LL << i)) cnt[i]++, ans++, p = max(i, p);
  if (k > n or ans > k) return puts("NO"), 0;
  while (ans < k and p >= 0) {
    while (!cnt[p] and p >= 0) p--;
    if (p < 0) break;
    cnt[p]--;
    cnt[p - 1] += 2;
    ans++;
  }
  if (ans < k) return puts("NO"), 0;
  puts("YES");
  for (int i = 0; i <= 32; i++)
    while (cnt[i]--) printf("%d ", (1 << i));
  puts("");
  return 0;
}