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 + 100;
const int maxpow = 1 << 17;
const int inf = 1e9 + 7;
long long h[maxn];
int par[maxn];
struct node {
  pair<int, int> a;
  node(pair<int, int> _a) : a(_a) {}
  node() : a(0, -1) {}
  void merge(node x, node y) { a = max(x.a, y.a); }
};
node tree[maxpow * 2];
node query(int i, int ns, int ne, int qs, int qe) {
  if (qe <= ns || qs >= ne) {
    return node();
  }
  if (ns >= qs && ne <= qe) {
    return tree[i];
  }
  int mid = (ns + ne) / 2;
  node res1 = query(2 * i, ns, mid, qs, qe);
  node res2 = query(2 * i + 1, mid, ne, qs, qe);
  node res;
  res.merge(res1, res2);
  return res;
}
void update(int i, int ns, int ne, int qs, pair<int, int> v) {
  if (qs + 1 <= ns || qs >= ne) {
    return;
  }
  if (ns + 1 == ne) {
    tree[i] = node(v);
    return;
  }
  int mid = (ns + ne) / 2;
  update(2 * i, ns, mid, qs, v);
  update(2 * i + 1, mid, ne, qs, v);
  tree[i].merge(tree[2 * i], tree[2 * i + 1]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, d;
  cin >> n >> d;
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    cin >> h[i];
    a.push_back(h[i]);
  }
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  pair<int, int> mx(-1, -1);
  for (int i = 0; i < n; i++) {
    int left = lower_bound(a.begin(), a.end(), h[i] - d + 1) - a.begin();
    int right = lower_bound(a.begin(), a.end(), h[i] + d) - a.begin();
    int cur = lower_bound(a.begin(), a.end(), h[i]) - a.begin();
    node ans1 = query(1, 0, maxpow, 0, left);
    node ans2 = query(1, 0, maxpow, right, n);
    pair<int, int> ans = max(ans1.a, ans2.a);
    ans.first += 1;
    par[i] = ans.second;
    ans.second = i;
    mx = max(mx, ans);
    update(1, 0, maxpow, cur, ans);
  }
  cout << mx.first << endl;
  int ind = mx.second;
  vector<int> res;
  while (ind != -1) {
    res.push_back(ind);
    ind = par[ind];
  }
  reverse(res.begin(), res.end());
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] + 1 << " ";
  }
  cout << endl;
}
