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>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B, T M) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M) {
  return BigMod(A, M - 2, M);
}
long long gcdr(long long a, long long b) {
  if (a == 0) return b;
  return gcdr(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void FastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
}
const long long size3 = 300005;
const long long size2 = 200005;
const long long size1 = 100005;
int main() {
  long long t, i, j, m, x, y, n;
  t = 1;
  while (t--) {
    scanf("%lld", &n);
    scanf("%lld", &m);
    vector<long long> listt;
    long long cnt = 0;
    long long k = m;
    for (i = 0; i < 31; i++) {
      if ((n & (1 << i)) != 0) {
        cnt++;
        listt.push_back((long long)(1 << i));
      }
    }
    if (cnt > m || n < m) {
      printf("NO");
      printf("\n");
      continue;
    }
    if (cnt == m) {
      printf("YES");
      printf("\n");
      for (i = 0; i < listt.size(); i++) {
        printf("%lld", listt[i]);
        printf(" ");
      }
      printf("\n");
      continue;
    }
    bool done = false;
    vector<long long> ans;
    queue<long long> Q;
    for (i = 0; i < listt.size() && cnt < m && !done; i++) {
      long long v = listt[i];
      if (v == 1) {
        ans.push_back(listt[i]);
        continue;
      } else {
        Q.push(v / 2);
        cnt++;
        v = v / 2;
        if (cnt == m) {
          ans.push_back(v);
          done = true;
        }
        while (!Q.empty() && cnt < m && !done) {
          if (v == 1) {
            ans.push_back(v);
            v = Q.front();
            Q.pop();
            if (Q.empty() && v == 1) {
              ans.push_back(1);
            }
          } else {
            v = v / 2;
            cnt++;
            Q.push(v);
            if (cnt == m) {
              ans.push_back(v);
              done = true;
            }
          }
        }
      }
    }
    for (j = i; j < listt.size(); j++) {
      ans.push_back(listt[j]);
    }
    while (!Q.empty()) {
      ans.push_back(Q.front());
      Q.pop();
    }
    printf("YES");
    printf("\n");
    for (i = 0; i < ans.size(); i++) {
      printf("%lld", ans[i]);
      printf(" ");
    }
    printf("\n");
  }
  return 0;
}