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;
template <typename T>
T read() {
  T p = 1, x = 0;
  char s = getchar();
  while (s == ' ' || s == '\n') s = getchar();
  if (s == '-') p = -1, s = getchar();
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * p;
}
template <typename A, typename B>
void Umax(A &a, const B &b) {
  if (a < b) a = b;
}
template <typename A, typename B>
void Umin(A &a, const B &b) {
  if (a > b) {
    a = b;
  }
}
long long bin_pow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 1) {
    long long cnt = a * bin_pow(a, n - 1);
    return cnt;
  } else {
    long long cnt = bin_pow(a, n / 2);
    return cnt * cnt;
  }
}
const int N = (int)1e6 + 10;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 10;
const long long LLINF = (long long)1e18 + 10;
long long check(const vector<int> &v) {
  long long s = 0;
  for (int i = 0; i < v.size(); ++i) {
    s += v[i];
  }
  return s;
}
void go(vector<int> v) {
  cout << "YES" << endl;
  reverse(v.begin(), v.end());
  for (long long i = 0; i < v.size(); ++i) {
    for (int j = 0; j < v[i]; ++j) cout << bin_pow(2LL, i) << ' ';
  }
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  if (n == k) {
    cout << "YES" << endl;
    for (int i = 0; i < n; ++i) {
      cout << 1 << ' ';
    }
    return 0;
  }
  vector<int> bits;
  long long zz = n;
  do {
    bits.push_back(zz % 2);
    zz /= 2;
  } while (zz);
  reverse(bits.begin(), bits.end());
  for (int i = 0; i < bits.size() - 1; ++i) {
    long long S = check(bits);
    if (k == S) {
      go(bits);
    } else {
      int z = bits[i];
      if (k >= S + z) {
        bits[i + 1] += z * 2;
        bits[i] = 0;
      } else {
        for (int j = 0; j < z; ++j) {
          bits[i + 1] += 2;
          bits[i]--;
          S++;
          if (S == k) {
            go(bits);
          }
        }
      }
    }
  }
  cout << "NO" << endl;
  return 0;
}