Marmot found a row with n pillars. The i-th pillar has the height of hi meters. Starting from one pillar i1, Marmot wants to jump on the pillars i2, ..., ik. (1 ≤ i1 < i2 < ... < ik ≤ n). From a pillar i Marmot can jump on a pillar j only if i < j and |hi - hj| ≥ d, where |x| is the absolute value of the number x.

Now Marmot is asking you find out a jump sequence with maximal length and print it.

Input

The first line contains two integers n and d (1 ≤ n ≤ 105, 0 ≤ d ≤ 109).

The second line contains n numbers h1, h2, ..., hn (1 ≤ hi ≤ 1015).

Output

The first line should contain one integer k, the maximal length of a jump sequence.

The second line should contain k integers i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), representing the pillars' indices from the maximal length jump sequence.

If there is more than one maximal length jump sequence, print any.

Examples

Input

5 2
1 3 6 7 4


Output

4
1 2 3 5 


Input

10 3
2 1 3 6 9 11 7 3 20 18


Output

6
1 4 6 7 8 9 

Note

In the first example Marmot chooses the pillars 1, 2, 3, 5 with the heights 1, 3, 6, 4. Another jump sequence of length 4 is 1, 2, 4, 5.
#include <bits/stdc++.h>
using namespace std;
long long n, d, a[100005], mx = 0, p = 0;
pair<long long, long long> first[100005];
vector<long long> v;
int main() {
  cin >> n >> d;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  first[1] = make_pair(1, -1);
  for (int i = 2; i <= n; ++i) {
    first[i] = make_pair(1, -1);
    for (int j = i - 1; j > 0 && i - j < 700; --j)
      if (fabs(a[j] - a[i]) >= d && first[j].first + 1 > first[i].first)
        first[i].first = first[j].first + 1, first[i].second = j;
  }
  for (int i = 1; i <= n; ++i)
    if (first[i].first > mx) mx = first[i].first, p = i;
  cout << first[p].first << endl;
  v.push_back(p);
  while (first[p].second != -1) {
    v.push_back(first[p].second);
    p = first[p].second;
  }
  for (int i = (int)v.size() - 1; i > 0; --i) cout << v[i] << " ";
  cout << v[0] << endl;
  return 0;
}
