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 MOD_EXPO(long long b, long long p, long long m) {
  if (p == 0) return 1;
  long long ret = MOD_EXPO(b, p / 2, m) % m;
  ret = (ret * ret) % m;
  return ((p & 1) ? (ret * b) % m : ret % m);
}
long long POWER(long long N, long long K) {
  long long i, ans = 1;
  for (i = 1; i <= K; i++) ans *= N;
  return ans;
}
int SET(int N, int pos) { return (N | (1 << pos)); }
int RESET(int N, int pos) { return (N & !(1 << pos)); }
bool CHECK(int N, int pos) { return (N & (1 << pos)); }
int dx4[] = {1, -1, 0, 0};
int dy4[] = {0, 0, 1, -1};
int dx6[] = {0, 0, 1, -1, 0, 0};
int dy6[] = {1, -1, 0, 0, 0, 0};
int dz6[] = {0, 0, 0, 0, 1, -1};
int dx8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dy8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dkx8[] = {-1, 1, -1, 1, -2, -2, 2, 2};
int dky8[] = {2, 2, -2, -2, 1, -1, 1, -1};
int tc = 1;
const double eps = 1e-9;
const double pi = acos(-1.0);
const long long int mn = 30;
const long long int mx = 1e5;
const long long int mod = 1e9 + 7;
int n, k;
priority_queue<int> pq;
int main() {
  int i, x, sum;
  vector<int> numbers;
  cin >> n >> k;
  for (i = 30; i >= 0; i--) {
    if (CHECK(n, i)) pq.push((1 << i));
  }
  while (pq.size() < k) {
    x = pq.top();
    if (x > 1) {
      pq.pop();
      pq.push(x / 2);
      pq.push(x / 2);
    } else {
      break;
    }
  }
  sum = 0;
  while (!pq.empty()) {
    x = pq.top();
    pq.pop();
    sum += x;
    numbers.push_back(x);
  }
  if (sum == n && numbers.size() == k) {
    printf("YES\n");
    for (i = 0; i < k; i++) {
      printf("%d ", numbers[i]);
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
  return 0;
}