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 int N = 3e5 + 100, TOF = 2e4;
long long n, k, sum, cnt[N], d_cnt[N];
bool mark[N];
vector<int> vec[N];
set<pair<int, int> > st;
void del(int num) {
  st.erase(pair<int, int>(cnt[num], num));
  mark[num] = true;
  for (int d : vec[num])
    if (mark[d] == false && d > 1) {
      st.erase(pair<int, int>(cnt[d], d));
      st.insert(pair<int, int>(--cnt[d], d));
    }
  for (int i = 2 * num; i <= n; i += num)
    if (mark[i] == false && i > 1) {
      st.erase(pair<int, int>(cnt[i], i));
      st.insert(pair<int, int>(--cnt[i], i));
    }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  for (int i = 1; i < N; i++)
    for (int j = 2 * i; j < N; j += i) d_cnt[j]++;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) sum += (n / i) - 1;
  if (sum < k) {
    cout << "No";
    return 0;
  }
  long long m, now = 0;
  for (int i = 1;; i++) {
    now += d_cnt[i];
    if (now >= k) {
      m = i;
      break;
    }
  }
  if (min(n, m) < TOF) {
    if (n > TOF) n = TOF;
    cerr << n << ' ' << k << '\n';
    for (int i = 1; i <= n; i++) {
      for (int j = 2 * i; j <= n; j += i) vec[j].push_back(i);
      if (i > 1) {
        cnt[i] = vec[i].size() + (n / i) - 1;
        st.insert(pair<int, int>(cnt[i], i));
      }
    }
    sum = 0;
    for (int i = 1; i <= n; i++) sum += d_cnt[i];
    k = sum - k;
    while (k) {
      pair<int, int> p = *(--st.lower_bound(pair<int, int>(k, N)));
      del(p.second);
      k -= p.first;
    }
    int CNT = 0;
    for (int i = 1; i <= n; i++)
      if (mark[i] == false) CNT++;
    cout << "Yes\n" << CNT << '\n';
    for (int num = 1; num <= n; num++)
      if (mark[num] == false) cout << num << ' ';
    return 0;
  }
  n = m;
  k -= now;
  for (int i = n; i && k < 0; i--)
    if (d_cnt[i] == 1) {
      k++;
      mark[i] = true;
    }
  int CNT = n;
  for (int i = 1; i <= n; i++) CNT -= mark[i];
  cout << "Yes\n" << CNT << '\n';
  for (int i = 1; i <= n; i++)
    if (mark[i] == false) cout << i << ' ';
  return 0;
}
