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;
long long ara[100010], soso[100010], d, pa[100010];
int tree[100010 * 4], n;
void update(int idx, int b, int e, int p, int v) {
  if (p > e || p < b) return;
  if (b == e) {
    tree[idx] = v;
    return;
  }
  int m = (b + e) >> 1;
  update((idx << 1), b, m, p, v);
  update(((idx << 1) + 1), m + 1, e, p, v);
  tree[idx] = max(tree[(idx << 1)], tree[((idx << 1) + 1)]);
}
int query(int idx, int b, int e, int l, int r) {
  if (b > r || e < l) return 0;
  if (l <= b && e <= r) return tree[idx];
  int m = (b + e) >> 1;
  return max(query((idx << 1), b, m, l, r),
             query(((idx << 1) + 1), m + 1, e, l, r));
}
int getId(long long v) { return (lower_bound(soso, soso + n, v) - soso); }
int main() {
  ios::sync_with_stdio(false);
  long long prin = 0, pos, a, b;
  cin >> n >> d;
  for (int i = 0; i <= n - 1; i++) cin >> ara[i], soso[i] = ara[i];
  sort(soso, soso + n);
  for (int i = 0; i <= n - 1; i++) {
    int low = getId(ara[i] - d), up = getId(ara[i] + d);
    if (ara[i] - d != soso[low]) low--;
    a = low;
    b = up;
    low = query(1, 0, 100010 - 1, 0, low) + 1;
    up = query(1, 0, 100010 - 1, up, 100010 - 1) + 1;
    pa[i] = max(low, up);
    update(1, 0, 100010 - 1, getId(ara[i]), pa[i]);
  }
  prin = tree[1];
  pos = -d;
  cout << prin << endl;
  stack<int> show;
  for (int i = n - 1; i >= 0; i--) {
    if (pa[i] == prin && abs(ara[i] - pos) >= d) {
      show.push(i + 1);
      pos = ara[i];
      prin--;
    }
  }
  while (!show.empty()) {
    cout << show.top() << ' ';
    show.pop();
  }
  return 0;
}
