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 typ1 {
  int v, pre, pos;
  typ1(int v = 0, int pre = 0, int pos = 0) : v(v), pre(pre), pos(pos) {}
};
typ1 t[2 * 1234567];
int f[1234567], pre[1234567];
int n, cnt;
long long a[1234567], h[1234567], d;
typ1 max(typ1 a, typ1 b) {
  if (a.v > b.v) return a;
  return b;
}
typ1 query(int v, int l, int r, int ql, int qr) {
  if (qr < l || r < ql) return typ1(0, 0, 0);
  if (ql <= l && r <= qr) return t[v];
  int midn = (l + r) >> 1, ls = v << 1, rs = ls + 1;
  typ1 a = query(ls, l, midn, ql, qr), b = query(rs, midn + 1, r, ql, qr);
  if (a.v < b.v)
    return b;
  else
    return a;
}
void update(int v, int l, int r, int ind, typ1 p) {
  if (l == r && l == ind) {
    if (t[v].v < p.v) t[v] = p;
    return;
  }
  int midn = (l + r) >> 1, ls = v << 1, rs = ls + 1;
  if (midn < ind)
    update(rs, midn + 1, r, ind, p);
  else
    update(ls, l, midn, ind, p);
  if (t[ls].v > t[rs].v)
    t[v] = t[ls];
  else
    t[v] = t[rs];
}
int l_x(long long x) {
  int l = 1, r = cnt;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (a[mid] <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return r;
}
int r_x(long long x) {
  int l = 1, r = cnt;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (a[mid] >= x)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
void init() {
  scanf("%d%I64d", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &h[i]);
    a[i] = h[i];
  }
  sort(a + 1, a + n + 1);
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i - 1] != a[i]) cnt++;
    a[cnt] = a[i];
  }
}
void print(int x) {
  if (x == 0) return;
  print(pre[x]);
  printf("%d ", x);
}
void solve() {
  int ans = 0, ids;
  for (int i = 1; i <= n; i++) {
    int l = l_x(h[i] - d), r = r_x(h[i] + d);
    if (a[l] > h[i] - d) l--;
    if (a[r] < h[i] + d) r++;
    typ1 p = max(query(1, 1, cnt, 1, l), query(1, 1, cnt, r, cnt));
    f[i] = ++p.v;
    pre[i] = p.pos;
    p.pre = p.pos;
    p.pos = i;
    if (f[i] > ans) {
      ans = f[i];
      ids = i;
    }
    update(1, 1, cnt, l_x(h[i]), p);
  }
  printf("%d\n", ans);
  print(ids);
}
int main() {
  init();
  solve();
}
