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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
inline void take_input(long long A[], long long n) {
  for (long long i = 0; i < n; i++) {
    cin >> A[i];
  }
}
inline void show(long long A[], long long n) {
  for (long long i = 0; i < n; i++) {
    cout << A[i] << " ";
  }
  cout << '\n';
}
inline void show_vect(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << '\n';
}
inline long long countDivisibles(long long a, long long b, long long c) {
  if (a % c == 0) {
    return (a % c + b % c + 1);
  }
  return (a % c + b % c);
}
inline bool sortbysec(const pair<long long, long long> &a,
                      const pair<long long, long long> &b) {
  return (a.second < b.second);
}
inline bool sortbysecdesc(const pair<long long, long long> &a,
                          const pair<long long, long long> &b) {
  return (a.second > b.second);
}
inline void swap(long long &a, long long &b) {
  long long temp = a;
  a = b;
  b = temp;
}
inline void initialise(long long A[], long long n, long long k) {
  for (long long i = 0; i < n; i++) {
    A[i] = k;
  }
}
inline void vect_input(vector<long long> &v, long long n) {
  for (long long i = 0; i < n; i++) {
    long long input;
    cin >> input;
    v.push_back(input);
  }
}
void p_g_t() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int32_t main() {
  p_g_t();
  long long n, k;
  cin >> n >> k;
  long long countof2s = __builtin_popcountll(n);
  if (countof2s > k) {
    cout << "NO";
  } else if (n == k) {
    cout << "YES" << '\n';
    for (long long i = 0; i < k / 2; i++) cout << 1 << " " << 1 << " ";
    if (k % 2 != 0) cout << 1;
  } else if (n == 1) {
    if (k > 1)
      cout << "NO";
    else {
      cout << "YES" << '\n' << 1;
    }
  } else {
    vector<long long> binary;
    long long flag = n;
    while (flag > 0) {
      binary.push_back(flag % 2);
      flag /= 2;
    }
    if (countof2s == k) {
      cout << "YES" << '\n';
      for (long long i = 0; i < binary.size(); i++) {
        if (binary[i] == 1) {
          cout << (long long)pow(2, i) << " ";
        }
      }
    } else {
      long long *dp = new long long[32];
      ;
      initialise(dp, 32, 0);
      for (long long i = 0; i < binary.size(); i++) {
        if (binary[i] == 1) {
          dp[i]++;
        }
      }
      long long extra = k - countof2s;
      while (extra--) {
        long long check;
        for (long long i = 31; i >= 1; i--) {
          if (dp[i] > 0) check = i;
        }
        dp[check]--;
        dp[check - 1]++;
        dp[check - 1]++;
      }
      bool check = true;
      for (long long i = 0; i < 32; i++) {
        if (dp[i] < 0) check = false;
      }
      if (!check)
        cout << "NO";
      else {
        cout << "YES" << '\n';
        for (long long i = 0; i < 32; i++) {
          if (dp[i] > 0) {
            long long y = (long long)pow(2, i);
            for (long long j = 0; j < dp[i]; j++) {
              cout << y << " ";
            }
          }
        }
      }
    }
  }
  return 0;
}