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 MAXN = 1000 * 100 + 10;
long long n, d, h[MAXN], c[MAXN];
int par[MAXN];
pair<int, int> seg[4 * MAXN];
pair<long long, int> a[MAXN];
pair<int, int> b[MAXN];
void add(int ind, int p, int val, int v, int second, int e) {
  if (e - second <= 1) {
    seg[v] = make_pair(val, ind);
    return;
  }
  int m = (second + e) / 2;
  if (p < m)
    add(ind, p, val, v * 2, second, m);
  else
    add(ind, p, val, v * 2 + 1, m, e);
  if (seg[v * 2].first > seg[v * 2 + 1].first)
    seg[v] = seg[v * 2];
  else
    seg[v] = seg[v * 2 + 1];
}
pair<int, int> get(int l, int r, int v, int second, int e) {
  if (e - second <= 1 || (l <= second && e <= r)) return seg[v];
  int m = (second + e) / 2;
  if (r <= m) return get(l, r, v * 2, second, m);
  if (m <= l) return get(l, r, v * 2 + 1, m, e);
  pair<int, int> a = get(l, r, v * 2, second, m);
  pair<int, int> b = get(l, r, v * 2 + 1, m, e);
  if (a.first > b.first)
    return a;
  else
    return b;
}
int main() {
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    cin >> h[i];
    a[i] = make_pair(h[i], i);
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    b[i] = make_pair(a[i].second, i);
    c[i] = a[i].first;
  }
  sort(b, b + n);
  int ans = 0, x = 0;
  for (int i = 0; i < n; i++) {
    int k1 = upper_bound(c, c + n, h[i] - d) - c;
    int k2 = lower_bound(c, c + n, h[i] + d) - c;
    pair<int, int> t1 = make_pair(0, -1), t2 = make_pair(0, -1);
    if (k1 > 0) t1 = get(0, max(0, k1), 1, 0, n);
    if (k2 < n) t2 = get(max(0, k2), n, 1, 0, n);
    if (t1.first < t2.first) swap(t1, t2);
    add(i, b[i].second, t1.first + 1, 1, 0, n);
    par[i] = t1.second;
    if (t1.first == 0) par[i] = -1;
    if (ans < t1.first + 1) ans = t1.first + 1, x = i;
  }
  cout << ans << endl;
  vector<int> vec;
  while (x != -1) {
    vec.push_back(x + 1);
    x = par[x];
  }
  for (int i = vec.size() - 1; i >= 0; i--) cout << vec[i] << " ";
  cout << endl;
}
