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 fenwik_tree {
  int mx[100100], mx_id[100100], ID;
  void add(int x, int val, int id) {
    while (x < 100100) {
      if (mx[x] < val) mx[x] = val, mx_id[x] = id;
      x += x & -x;
    }
  }
  int calc(int x) {
    ID = 0;
    int rlt = 0;
    while (x) {
      if (rlt < mx[x]) rlt = mx[x], ID = mx_id[x];
      x -= x & -x;
    }
    return rlt;
  }
} P, S;
int n, d, a[100100], prv[100100], nxt[100100], f[100100], bf[100100];
long long h[100100], x[100100];
map<long long, int> mp;
void print(int x) {
  if (!x) return;
  print(bf[x]);
  printf("%d ", x);
}
int main() {
  scanf("%d %d", &n, &d);
  for (int i = 1; i <= n; i++) scanf("%I64d", &h[i]), x[i - 1] = h[i];
  sort(x, x + n);
  int m = unique(x, x + n) - x;
  for (int i = 0; i < m; i++) mp[x[i]] = i + 1;
  for (int i = 1; i <= n; i++) {
    a[i] = mp[h[i]];
    prv[i] = lower_bound(x, x + m, h[i] - d + 1) - x;
    nxt[i] = lower_bound(x, x + m, h[i] + d) - x + 1;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    f[i] = P.calc(prv[i]) + 1, bf[i] = P.ID;
    int rlt = S.calc(100100 - nxt[i]) + 1;
    if (f[i] < rlt) f[i] = rlt, bf[i] = S.ID;
    P.add(a[i], f[i], i);
    S.add(100100 - a[i], f[i], i);
    ans = max(ans, f[i]);
  }
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++) {
    if (f[i] == ans) {
      print(i);
      puts("");
      return 0;
    }
  }
  return 0;
}
