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;
const int N = 1000 * 100 + 10;
int n, d[N], par[N];
pair<int, int> tr[N * 4];
pair<long long, int> a[N], b[N];
void add(int val, int p, int b = 0, int e = n, int ind = 1) {
  tr[ind] = max(tr[ind], make_pair(val, p));
  if (b + 1 == e) return;
  int mid = (b + e) / 2;
  if (p < mid)
    add(val, p, b, mid, ind * 2);
  else
    add(val, p, mid, e, ind * 2 + 1);
  return;
}
pair<int, int> get(int l, int r, int b = 0, int e = n, int ind = 1) {
  if (l <= b && e <= r) return tr[ind];
  int mid = (b + e) / 2;
  pair<int, int> ret = make_pair(0, 0);
  if (l < mid) ret = get(l, r, b, mid, ind * 2);
  if (r > mid) ret = max(get(l, r, mid, e, ind * 2 + 1), ret);
  return ret;
}
void pp(int v) {
  if (d[v] == 1) {
    cout << v + 1 << " ";
    return;
  }
  pp(par[v]);
  cout << v + 1 << " ";
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    b[i].first = a[i].first;
    b[i].second = i;
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) a[b[i].second].second = i;
  d[0] = 1;
  add(1, a[0].second);
  for (int i = 1; i < n; i++) {
    int l = upper_bound(b, b + n, make_pair(a[i].first - k, n)) - b;
    int r = lower_bound(b, b + n, make_pair(a[i].first + k, 0)) - b;
    if (l) {
      pair<int, int> temp = get(0, l);
      d[i] = temp.first + 1;
      par[i] = b[temp.second].second;
    }
    if (r < n) {
      pair<int, int> temp = get(r, n);
      if (temp.first >= d[i]) {
        d[i] = temp.first + 1;
        par[i] = b[temp.second].second;
      }
    }
    add(d[i], a[i].second);
  }
  long long maxx = 0, maxind;
  for (int i = 0; i < n; i++) {
    if (d[i] > maxx) {
      maxx = d[i];
      maxind = i;
    }
  }
  cout << maxx << endl;
  pp(maxind);
}
