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;
inline long long read() {
  long long num = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num * neg;
}
long long deg[300010], ans, n, k, tmp[300010];
bool cmp1(long long x, long long y) { return deg[x] > deg[y]; }
inline void solve() {
  for (long long i = 1; i <= n; i++)
    for (long long j = 2; i * j <= n; j++) deg[i * j]++;
  for (long long i = 1; i <= n; i++) ans += deg[i];
  if (ans < k) {
    printf("No\n");
    return;
  }
  if (k == 0) {
    printf("Yes\n");
    printf("1\n1");
    return;
  }
  long long m = n, cnt = 0;
  while (ans - deg[m] >= k) ans -= deg[m], m--;
  set<long long> res;
  for (long long i = m; i > m / 2; i--) tmp[++cnt] = i;
  for (long long i = 1; i <= m; i++) res.insert(i);
  sort(tmp + 1, tmp + cnt + 1, cmp1);
  for (long long i = 1; i <= cnt; i++) {
    if (ans - deg[tmp[i]] >= k) {
      ans -= deg[tmp[i]], res.erase(tmp[i]);
    }
  }
  set<long long>::iterator it = res.begin();
  if (ans == k) {
    printf("Yes\n");
    printf("%d\n", res.size());
    for (; it != res.end(); ++it) printf("%lld ", *it);
  } else
    printf("No\n");
}
signed main() {
  n = read(), k = read();
  solve();
  return 0;
}
