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;
int length[200001];
struct node {
  int l, r, mid;
  int longest;
  long long int big, small;
  node *lch, *rch;
  node(int a, int b) {
    l = a;
    r = b;
    longest = 0;
    big = -1e15;
    small = 1e15;
    mid = (a + b) / 2;
    lch = rch = 0;
  }
  void update(int val, long long int h, int ord) {
    if (h > big) big = h;
    if (h < small) small = h;
    if (length[longest] < length[ord]) longest = ord;
    if (l == r) return;
    if (val <= mid) {
      if (lch == 0) lch = new node(l, mid);
      lch->update(val, h, ord);
    } else {
      if (rch == 0) rch = new node(mid + 1, r);
      rch->update(val, h, ord);
    }
  }
  int getans(int val, long long int h, long long int d) {
    if (small - h >= d || h - big >= d) return longest;
    int a = 0, b = 0;
    if (lch != 0) {
      if (h - lch->small >= d || lch->big - h >= d) {
        a = lch->getans(val, h, d);
      }
    }
    if (rch != 0) {
      if (rch->big - h >= d || h - rch->small >= d) {
        b = rch->getans(val, h, d);
      }
    }
    return (length[a] >= length[b]) ? a : b;
  }
};
pair<pair<long long int, int>, int> arr[200001];
int pre[200001];
vector<int> v[200001];
bool cmp(pair<pair<long long int, int>, int> p,
         pair<pair<long long int, int>, int> q) {
  return p.first.second < q.first.second;
}
vector<int> outs;
int main() {
  int n;
  long long int d;
  cin >> n >> d;
  node *root = new node(1, n);
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i].first.first;
    arr[i].first.second = i;
  }
  sort(arr + 1, arr + n + 1);
  for (int i = 1; i <= n; ++i) {
    arr[i].second = i;
  }
  sort(arr + 1, arr + n + 1, cmp);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    int g = root->getans(arr[i].second, arr[i].first.first, d);
    length[i] = length[g] + 1;
    v[g].emplace_back(i);
    pre[i] = g;
    if (length[i] > length[ans]) ans = i;
    root->update(arr[i].second, arr[i].first.first, i);
  }
  vector<int> out;
  cout << length[ans] << '\n';
  while (ans != 0) {
    out.emplace_back(ans);
    ans = pre[ans];
  }
  cout << out[out.size() - 1];
  for (int i = out.size() - 2; i >= 0; --i) {
    cout << ' ' << out[i];
  }
  cout << endl;
  return 0;
}
