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 = 1e5 + 10;
int dp[MAXN], sz, n, par[MAXN];
long long a[MAXN], b[MAXN];
map<long long, int> mp;
pair<long long, int> data[4 * MAXN];
void update(int p, int x, int pos, int s, int e, int id) {
  if (data[id].first < x) {
    data[id].first = x;
    data[id].second = pos;
  }
  if ((e - s) < 2) {
    return;
  }
  int mid = (s + e) / 2;
  if (p < mid) {
    update(p, x, pos, s, mid, 2 * id);
  } else {
    update(p, x, pos, mid, e, 2 * id + 1);
  }
}
pair<long long, int> get(int l, int r, int s, int e, int id) {
  if (e <= l || r <= s) {
    return make_pair(-1, 0);
  }
  if (l <= s && e <= r) {
    return data[id];
  }
  int mid = (s + e) / 2;
  pair<long long, int> ans = get(l, r, s, mid, 2 * id);
  ans = max(get(l, r, mid, e, 2 * id + 1), ans);
  return ans;
}
int main() {
  for (int i = 0; i < MAXN; i++) par[i] = -1;
  int d;
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    mp[b[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int r, l;
    r = upper_bound(b, b + n, a[i] - d) - b;
    l = lower_bound(b, b + n, a[i] + d) - b;
    int mx = -1, idx = 0;
    if (a[i] - d >= 0 && r > 0) {
      pair<int, int> cur = get(0, r, 0, n, 1);
      mx = cur.first;
      idx = cur.second;
    }
    if (l != n) {
      pair<int, int> cur = get(l, n, 0, n, 1);
      if (cur.first > mx) {
        mx = cur.first;
        idx = cur.second;
      }
    }
    dp[i] = mx + 1;
    par[i] = idx;
    update(mp[a[i]], dp[i], i, 0, n, 1);
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    if (dp[i] > dp[ans]) ans = i;
  cout << dp[ans] << endl;
  int cur = ans;
  vector<int> v;
  for (int i = 0; i < dp[ans]; i++) {
    v.push_back(cur);
    cur = par[cur];
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) cout << v[i] + 1 << " ";
  cout << endl;
  return 0;
}
