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;
pair<long long, long long> p[100004];
long long n, d, c;
map<long long, long long> m;
struct node {
  int val, idx;
} st[4000004];
void Update(int node, int l, int r, int i, int val, int idx) {
  if (i > r || i < l) return;
  if (i == l && i == r) {
    st[node].val = val;
    st[node].idx = idx;
    return;
  }
  int mid = (l + r) / 2;
  Update(node * 2, l, mid, i, val, idx);
  Update(node * 2 + 1, mid + 1, r, i, val, idx);
  st[node] = (st[node * 2 + 1].val >= st[node * 2].val) ? st[node * 2 + 1]
                                                        : st[node * 2];
}
node Query(int i, int j, int p = 1, int l = 0, int r = c - 1) {
  if (i > r || j < l) return node();
  if (i <= l && j >= r) return st[p];
  int mid = (l + r) / 2;
  node lf = Query(i, j, p * 2, l, mid);
  node rt = Query(i, j, p * 2 + 1, mid + 1, r);
  return (lf.val >= rt.val) ? lf : rt;
}
int par[100004];
int main() {
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    m[x], m[x + d], m[x - d];
    p[i] = {x, i + 1};
  }
  for (map<long long, long long>::iterator it = m.begin(); it != m.end(); it++)
    it->second = c++;
  for (int i = n - 1; i >= 0; i--) {
    node x = Query(0, m[p[i].first - d]);
    node y = Query(m[p[i].first + d], c - 1);
    if (x.val > y.val) {
      par[p[i].second] = x.idx;
      Update(1, 0, c - 1, m[p[i].first], x.val + 1, p[i].second);
    } else {
      par[p[i].second] = y.idx;
      Update(1, 0, c - 1, m[p[i].first], y.val + 1, p[i].second);
    }
  }
  cout << st[1].val << endl;
  int cur = st[1].idx;
  while (true) {
    if (!cur) break;
    cout << cur << " ";
    cur = par[cur];
  }
  return 0;
}
