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;
class node {
 public:
  int jumps;
  long long currentPillar;
  int index;
  node *previous;
  node(int _jumps, long long _currentPillar, int _index, node *_previous) {
    jumps = _jumps;
    currentPillar = _currentPillar;
    index = _index;
    previous = _previous;
  }
};
typedef node *Node;
Node deadEnd = new node(0, -1, -1, NULL);
vector<Node> tree;
void update(int a, Node v, int i, int start, int end) {
  if (end - start == 1) {
    tree[i] = v;
    return;
  }
  int mid = (start + end) / 2;
  if (a < mid)
    update(a, v, i * 2, start, mid);
  else
    update(a, v, i * 2 + 1, mid, end);
  if (tree[i * 2]->jumps > tree[i * 2 + 1]->jumps)
    tree[i] = tree[i * 2];
  else
    tree[i] = tree[i * 2 + 1];
}
void printNode(Node n) {
  while (n != deadEnd) {
    if (n == n->previous)
      cout << "-------------ffs.... " << n << " " << n->previous << endl;
    cout << n->currentPillar << " ";
    n = n->previous;
  }
  cout << "deadEnd" << endl;
}
Node query(int a, int b, int i, int start, int end) {
  if (i == 0) cout << "FFS";
  if (start == a && end == b) return tree[i];
  int mid = (start + end) / 2;
  Node left = deadEnd, right = deadEnd;
  if (a < mid) left = query(a, min(b, mid), i * 2, start, mid);
  if (b > mid) right = query(max(a, mid), b, i * 2 + 1, mid, end);
  if (left->jumps > right->jumps)
    return left;
  else
    return right;
}
int main() {
  int n;
  long long d;
  cin >> n >> d;
  tree = vector<Node>(n << 2, deadEnd);
  vector<long long> pillars(n);
  for (int i = 0; i < n; i++) cin >> pillars[i];
  set<long long> pillarSet;
  for (int i = 0; i < n; i++) pillarSet.insert(pillars[i]);
  int maxIndex = pillarSet.size();
  map<long long, int> pillarToIndex;
  int index = 0;
  for (long long pillar : pillarSet) {
    pillarToIndex[pillar] = index;
    index++;
  }
  for (int i = pillars.size() - 1; i >= 0; i--) {
    long long pillar = pillars[i];
    Node best = deadEnd, temp;
    long long lower = pillar - d;
    auto nextLower = pillarSet.upper_bound(lower);
    if (*nextLower > lower && nextLower != pillarSet.begin())
      nextLower = prev(nextLower);
    if (*nextLower <= lower) {
      temp = query(0, pillarToIndex[*nextLower] + 1, 1, 0, maxIndex);
      if (temp->jumps > best->jumps) best = temp;
    }
    long long upper = pillar + d;
    auto nextUpper = pillarSet.lower_bound(upper);
    if (*nextUpper < upper && nextUpper != pillarSet.end())
      nextUpper = next(nextUpper);
    if (*nextUpper >= upper && nextUpper != pillarSet.end()) {
      temp = query(pillarToIndex[*nextUpper], maxIndex, 1, 0, maxIndex);
      if (temp->jumps > best->jumps) best = temp;
    }
    Node toAdd = new node(best->jumps + 1, pillar, i + 1, best);
    update(pillarToIndex[pillar], toAdd, 1, 0, maxIndex);
  }
  Node best = query(0, maxIndex, 1, 0, maxIndex);
  cout << best->jumps << endl;
  while (best != deadEnd) {
    cout << best->index << " ";
    best = best->previous;
  }
  return 0;
}
