Imp is really pleased that you helped him. But it you solve the last problem, his gladness would raise even more.

<image> Let's define <image> for some set of integers <image> as the number of pairs a, b in <image>, such that:

  * a is strictly less than b; 
  * a divides b without a remainder. 



You are to find such a set <image>, which is a subset of {1, 2, ..., n} (the set that contains all positive integers not greater than n), that <image>. 

Input

The only line contains two integers n and k <image>.

Output

If there is no answer, print "No".

Otherwise, in the first line print "Yes", in the second — an integer m that denotes the size of the set <image> you have found, in the second line print m integers — the elements of the set <image>, in any order.

If there are multiple answers, print any of them.

Examples

Input

3 3


Output

No


Input

6 6


Output

Yes
5
1 2 4 5 6 

Input

8 3


Output

Yes
4
2 4 5 8

Note

In the second sample, the valid pairs in the output set are (1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6). Thus, <image>.

In the third example, the valid pairs in the output set are (2, 4), (4, 8), (2, 8). Thus, <image>.
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (long long)3e5 + 5;
long long a[MAXN];
long long cnt[MAXN];
long long e[MAXN];
set<long long> bad_m = {6,  10, 12, 16, 18, 24, 28, 30,
                        36, 40, 42, 48, 60, 72, 96, 120};
void precalc() {
  for (long long i = 1; i < MAXN; i++) {
    for (long long j = i + i; j < MAXN; j += i) {
      cnt[j]++;
    }
  }
  e[1] = 0;
  for (long long i = 2; i < MAXN; i++) {
    e[i] = e[i - 1] + cnt[i];
  }
}
long long solve(long long n, long long k) {
  long long pos = -1;
  for (long long i = 1; i <= n; i++) {
    if (e[i] >= k) {
      pos = i;
      break;
    }
  }
  set<long long> numbers;
  for (long long i = 1; i <= pos; i++) {
    numbers.insert(i);
  }
  long long cur = e[pos];
  for (long long i = pos / 2 + 1; i <= pos; i++) {
    if (cur > k && cnt[i] == 1) {
      cur--;
      numbers.erase(i);
    }
  }
  cout << "Yes" << endl;
  cout << (long long)numbers.size() << endl;
  for (long long el : numbers) {
    cout << el << " ";
  }
  cout << endl;
  exit(0);
}
long long brute_force(long long n, long long k) {
  long long pos = -1;
  for (long long i = 1; i <= n; i++) {
    if (e[i] >= k) {
      pos = i;
      break;
    }
  }
  long long exceed = e[pos] - k;
  vector<long long> dp(exceed + 5, 0);
  vector<long long> par(exceed + 5, 0);
  dp[0] = 1;
  for (long long x = pos / 2 + 1; x <= pos; x++) {
    for (long long i = exceed; i >= 0; i--) {
      if (dp[i] && i + cnt[x] <= exceed && dp[i + cnt[x]] == 0) {
        dp[i + cnt[x]] = 1;
        par[i + cnt[x]] = x;
      }
    }
  }
  long long fn = exceed;
  set<long long> numbers;
  for (long long i = 1; i <= pos; i++) {
    numbers.insert(i);
  }
  while (fn != 0) {
    numbers.erase(par[fn]);
    fn -= cnt[par[fn]];
  }
  cout << "Yes" << endl;
  cout << (long long)numbers.size() << endl;
  for (long long el : numbers) {
    cout << el << " ";
  }
  exit(0);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  precalc();
  long long n, k;
  cin >> n >> k;
  long long pos = -1;
  for (long long i = 1; i <= n; i++) {
    if (e[i] >= k) {
      pos = i;
      break;
    }
  }
  if (pos == -1) {
    cout << "No" << endl;
    exit(0);
  }
  if (pos <= 120) {
    brute_force(n, k);
  } else {
    solve(n, k);
  }
}
