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 INF = (int)1E9;
int val[1 << 19], vid[1 << 19];
void add(int k, int nl, int nr, int x, pair<int, int> v) {
  if (x < nl || x > nr) return;
  if (nl == nr) {
    if (v.first > val[k]) {
      val[k] = v.first;
      vid[k] = v.second;
    }
    return;
  }
  int nm = (nl + nr) >> 1;
  add(2 * k + 1, nl, nm, x, v);
  add(2 * k + 2, nm + 1, nr, x, v);
  if (val[2 * k + 1] > val[2 * k + 2]) {
    val[k] = val[2 * k + 1];
    vid[k] = vid[2 * k + 1];
  } else {
    val[k] = val[2 * k + 2];
    vid[k] = vid[2 * k + 2];
  }
}
pair<int, int> get(int k, int nl, int nr, int l, int r) {
  if (r < nl || l > nr) return pair<int, int>(0, -1);
  if (l <= nl && nr <= r) return pair<int, int>(val[k], vid[k]);
  int nm = (nl + nr) >> 1;
  pair<int, int> a = get(2 * k + 1, nl, nm, l, r);
  pair<int, int> b = get(2 * k + 2, nm + 1, nr, l, r);
  if (a.first > b.first)
    return a;
  else
    return b;
}
int N, H;
int pre[100005];
long long D;
long long h[100005], hs[100005];
int aa[100005], A;
int main() {
  cin >> N >> D;
  for (int i = (0); i < (N); i++) {
    scanf("%I64d", h + i);
    hs[H++] = h[i];
  }
  sort(hs, hs + H);
  H = unique(hs, hs + H) - hs;
  hs[H] = (long long)1E18;
  memset(vid, -1, sizeof(vid));
  int ans = 0, e;
  for (int i = (0); i < (N); i++) {
    long long x = h[i];
    int hi = lower_bound(hs, hs + H, x) - hs;
    int pl = upper_bound(hs, hs + H, x - D) - hs - 1;
    int pr = lower_bound(hs, hs + H, x + D) - hs;
    pair<int, int> p = get(0, 0, H - 1, 0, pl);
    pair<int, int> p2 = get(0, 0, H - 1, pr, H - 1);
    if (p2.first > p.first) p = p2;
    pre[i] = p.second;
    int best = p.first + 1;
    add(0, 0, H - 1, hi, pair<int, int>(best, i));
    if (best > ans) {
      ans = best;
      e = i;
    }
  }
  printf("%d\n", ans);
  while (e != -1) {
    aa[A++] = e;
    e = pre[e];
  }
  reverse(aa, aa + A);
  for (int i = (0); i < (A); i++) printf("%d ", aa[i] + 1);
  return 0;
}
