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;
struct lsqy {
  int loc, id, l, r;
  long long mx;
} tr[100010 << 2];
long long a[100010], b[100010], hs[100010], hb[100010], f[1000100];
void build(int rt, int l, int r) {
  tr[rt].l = l;
  tr[rt].r = r;
  tr[rt].mx = 0;
  if (l == r) {
    return;
  }
  int mid = l + r >> 1;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
}
int query(int rt, int L, int R) {
  if (L <= tr[rt].l && tr[rt].r <= R) return tr[rt].loc;
  int locl = 0, locr = 0;
  int mid = tr[rt].l + tr[rt].r >> 1;
  if (L <= mid) locl = query(rt << 1, L, R);
  if (mid < R) locr = query(rt << 1 | 1, L, R);
  if (tr[locl].mx >= tr[locr].mx)
    return locl;
  else
    return locr;
}
void add(int rt, int pos, long long k) {
  if (pos == tr[rt].l && tr[rt].r == pos) {
    tr[rt].mx = k;
    tr[rt].loc = rt;
    return;
  }
  int mid = tr[rt].l + tr[rt].r >> 1;
  if (pos <= mid) add((rt << 1), pos, k);
  if (pos > mid) add((rt << 1 | 1), pos, k);
  tr[rt].mx = max(tr[(rt << 1)].mx, tr[(rt << 1 | 1)].mx);
  if (tr[(rt << 1)].mx >= tr[(rt << 1 | 1)].mx)
    tr[rt].loc = tr[(rt << 1)].loc;
  else
    tr[rt].loc = tr[(rt << 1 | 1)].loc;
}
int path[100010], rph[100010], id;
int main() {
  int n, d;
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    b[i] = a[i];
  }
  a[n + 1] = b[n + 1] = 1e15 + 100;
  sort(b + 1, b + 2 + n);
  int siz = unique(b + 1, b + 2 + n) - b - 1;
  for (int i = 1; i <= siz - 1; ++i) {
    hs[i] = upper_bound(b + 1, b + 1 + siz, b[i] - d) - b - 1;
    hb[i] = lower_bound(b + 1, b + 1 + siz, b[i] + d) - b;
  }
  build(1, 1, siz - 1);
  for (int i = 1; i <= n; ++i) {
    int sm = 0, bg = 0;
    a[i] = lower_bound(b + 1, b + 1 + siz, a[i]) - b;
    if (hs[a[i]]) sm = query(1, 1, hs[a[i]]);
    if (hb[a[i]] <= siz - 1) bg = query(1, hb[a[i]], siz - 1);
    if (tr[sm].mx >= tr[bg].mx) {
      path[i] = tr[sm].l;
      f[i] = tr[sm].mx + 1;
    } else {
      path[i] = tr[bg].l;
      f[i] = tr[bg].mx + 1;
    }
    add(1, a[i], f[i]);
    if (f[i] > f[id]) id = i;
  }
  printf("%d\n", f[id]);
  int pt = path[id];
  int tot = 0;
  rph[++tot] = id;
  for (int i = id - 1; i >= 1; i--) {
    if (a[i] == pt) {
      rph[++tot] = i;
      pt = path[i];
      if (pt == 0) break;
    }
  }
  for (int i = tot; i >= 1; i--) printf("%d ", rph[i]);
  return 0;
}
