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>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
int n, k;
int dp[300300];
vector<int> g[300300];
void solve() {
  cin >> n >> k;
  int ans = 0;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i * 2; j <= n; j += i) {
      g[j].push_back(i);
      ++ans;
    }
    mx = max(mx, (int)g[i].size());
  }
  int last = 2;
  int cnt = 0;
  for (int j = 1; j <= n; j++) {
    while (last * 2 <= j) {
      if (g[last].size() == 1) cnt--;
      last++;
    }
    if (g[j].size() == 1) cnt++;
    if (g[j].size() > cnt + 1) {
    }
  }
  if (ans < k) {
    cout << "No" << endl;
    return;
  }
  ans = 0;
  int t = 0;
  while (ans < k) {
    ans += g[t].size();
    t++;
  }
  t--;
  if (ans == k) {
    cout << "Yes\n";
    cout << t << "\n";
    for (int i = 1; i <= t; i++) {
      printf("%d ", i);
    }
    return;
  }
  for (int i = 1; i <= t; i++) {
    if (ans - k == g[i].size() + t / i - 1) {
      cout << "Yes\n";
      cout << t - 1 << "\n";
      for (int j = 1; j <= t; j++) {
        if (j != i) printf("%d ", j);
      }
      return;
    }
  }
  vector<int> D;
  for (int i = 1; i <= t; i++) {
    if (i + i > t && g[i].size() == 1) {
      D.push_back(i);
    }
  }
  vector<int> G(n * 2 + 10, 0);
  if (D.size() >= ans - k) {
    cout << "Yes\n";
    int dd = ans - k;
    for (int j = 1; j <= t; j++) G[j] = 1;
    for (int j = 0; j < dd; j++) G[D[j]] = 0;
    D.clear();
    for (int j = 1; j <= t; j++)
      if (G[j]) {
        D.push_back(j);
      }
    cout << D.size() << "\n";
    for (int j = 0; j < D.size(); j++) {
      printf("%d ", D[j]);
    }
    return;
  }
  if (t <= 20) {
    for (int i = 0; i < (1 << t); i++) {
      int cur = 0;
      for (int j = 0; j < t; j++) {
        if (i & (1 << j)) {
          for (int x = j + 1; x < t; x++) {
            if ((i & (1 << x)) && (x + 1) % (j + 1) == 0) {
              cur++;
            }
          }
        }
      }
      if (cur == k) {
        cout << "Yes\n";
        D.clear();
        for (int x = 0; x < t; x++) {
          if ((i & (1 << x))) {
            D.push_back(x + 1);
          }
        }
        cout << D.size() << "\n";
        for (int j = 0; j < D.size(); j++) {
          printf("%d ", D[j]);
        }
        return;
      }
    }
  }
  D.clear();
  for (int i = 1; i <= t; i++) {
    if (g[i].size() == 1) D.push_back(i);
  }
  int CC = 0;
  reverse(D.begin(), D.end());
  int pr = min(20, (int)D.size());
  for (int i = 0; i < (1 << pr); i++) {
    int cur = ans;
    for (int j = 0; j < pr; j++) {
      if (i & (1 << j)) cur -= t / D[j];
    }
    if (cur == k) {
      cout << "Yes\n";
      for (int x = 0; x < pr; x++) {
        if ((i & (1 << x))) {
          G[D[x]] = 1;
        }
      }
      D.clear();
      for (int j = 1; j <= t; j++)
        if (G[j] == 0) D.push_back(j);
      cout << D.size() << "\n";
      for (int j = 0; j < D.size(); j++) {
        printf("%d ", D[j]);
      }
      return;
    }
  }
  assert(false);
  cout << "No" << endl;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  char c;
  return 0;
}
