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;
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
int msb(long long int n) {
  int ans;
  for (int i = 0; i < 64; i++)
    if (n & (1LL << i)) ans = i + 1;
  return ans;
}
int SetBit(int n, int X) { return n | (1 << X); }
int ClearBit(int n, int X) { return n & ~(1 << X); }
int ToggleBit(int n, int X) { return n ^ (1 << X); }
bool CheckBit(int n, int X) { return (bool)(n & (1 << X)); }
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
bool isPerfectSquare(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long int lowerbound(long long int k, vector<long long int> v) {
  long long int l = 0;
  long long int size = v.size();
  long long int r = size - 1;
  long long int mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] < k)
      l = mid + 1;
    else {
      r = mid - 1;
    }
  }
  return l;
}
long long int upperbound(long long int k, vector<long long int> v) {
  long long int l = 0;
  long long int size = v.size();
  long long int r = size - 1;
  long long int mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] > k)
      r = mid - 1;
    else {
      l = mid + 1;
    }
  }
  return l;
}
long long int digits(long long int a) { return (floor(log10(a)) + 1); }
vector<long long int> primeFactors(long long int n) {
  vector<long long int> sol;
  bool once = false;
  while (n % 2 == 0) {
    if (!once) {
      sol.push_back(2);
      once = true;
    }
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    bool once = false;
    while (n % i == 0) {
      if (!once) {
        sol.push_back(i);
        once = true;
      }
      n = n / i;
    }
  }
  if (n > 2) {
    sol.push_back(n);
  }
  return (sol);
}
long long int fastPower(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  if (a == 0) return 0;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    b = b >> 1;
    a = (a * a) % 1000000007;
  }
  return res;
}
unordered_map<char, int> symbol = {{'[', -1}, {'(', -2}, {'{', -3},
                                   {']', 1},  {')', 2},  {'}', 3}};
bool isBalanced(string s) {
  stack<char> st;
  for (char bracket : s) {
    if (symbol[bracket] < 0) {
      st.push(bracket);
    } else {
      if (st.empty()) {
        return false;
      }
      char top = st.top();
      st.pop();
      if (symbol[top] + symbol[bracket] != 0) return false;
    }
  }
  if (st.empty())
    return true;
  else
    return false;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int C(long long int n, long long int k) {
  long long int ans = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 1; i <= k; i++) {
    ans = ans * ((n - i + 1));
    ans = ans / i;
  }
  return ans;
}
int setBits(int n) {
  int ctr = 0;
  while (n) {
    ctr++;
    n = n & (n - 1);
  }
  return ctr;
}
void printBinary(int n) {
  for (int i = 10; i >= 0; i--) {
    cout << ((n >> i) & 1);
  }
  cout << endl;
}
long long int ceil_div(long long int a, long long int b) {
  return (a + b - 1) / b;
}
long long int closest_strictly_greater_multiple(long long int n,
                                                long long int L) {
  return (L / n + 1) * n;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int count_digits(long long int n) { return floor(log10(n) + 1); }
void solve() {
  int n, k;
  cin >> n >> k;
  queue<int> q;
  map<int, int> ans;
  for (int i = 0; i <= 30; i++) {
    if ((n & (1 << i))) {
      if (i > 0) q.push(1 << i);
      ans[1 << i]++;
    }
  }
  if (k < (ans.size())) {
    cout << "NO" << endl;
    return;
  }
  int cnt = ans.size();
  while (cnt < k && !q.empty()) {
    int x = q.front();
    q.pop();
    ans[x]--;
    ans[x / 2] += 2;
    if (x / 2 > 1) {
      q.push(x / 2);
      q.push(x / 2);
    }
    cnt++;
  }
  if (cnt < k) {
    cout << "NO" << endl;
    return;
  }
  cout << "YES" << endl;
  for (auto it : ans) {
    int ctr = it.second;
    while (ctr--) {
      cout << it.first << " ";
    }
  }
  cout << endl;
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}