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, h[100005];
pair<long long, int> p[100005];
pair<int, int> ans[100005];
pair<int, int> t[400005];
void update(int l, int r, int pos, int x, int val) {
  if (l == r) {
    t[pos] = {val, l};
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(l, mid, pos << 1, x, val);
  else
    update(mid + 1, r, (pos << 1) + 1, x, val);
  t[pos] = max(t[pos << 1], t[(pos << 1) + 1]);
}
pair<int, int> query(int l, int r, int pos, int ql, int qr) {
  if (ql <= l && r <= qr) return t[pos];
  if (ql > r || qr < l) return {0, -1};
  int mid = (l + r) >> 1;
  return max(query(l, mid, pos << 1, ql, qr),
             query(mid + 1, r, (pos << 1) + 1, ql, qr));
}
void f(int i) {
  if (i == -1) return;
  f(ans[i].second);
  cout << i + 1 << ' ';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first;
    p[i].second = i;
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    h[p[i].second] = i;
  }
  for (int i = 0; i < n; i++) {
    int sm =
        upper_bound(p, p + n, make_pair(p[h[i]].first - d, (int)n)) - p - 1;
    int bg = lower_bound(p, p + n, make_pair(p[h[i]].first + d, 0)) - p;
    ans[i] = max(query(0, n - 1, 1, 0, sm), query(0, n - 1, 1, bg, n - 1));
    if (ans[i].first == 0)
      ans[i].second = -1;
    else
      ans[i].second = p[ans[i].second].second;
    ans[i].first++;
    update(0, n - 1, 1, h[i], ans[i].first);
  }
  int maxi = 0, pos = 0;
  for (int i = 0; i < n; i++) {
    if (ans[i].first > maxi) {
      maxi = ans[i].first;
      pos = i;
    }
  }
  cout << maxi << '\n';
  f(pos);
}
