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>
#pragma warning(disable : 4786)
#pragma warning(disable : 4996)
using namespace std;
pair<int, int> tree[4 * 100005];
long long int H[100005], h[100005];
int prevs[100005], dp[100005];
void insert(int at, int l, int r, long long int x, int v, int id) {
  if (l == r) {
    if (tree[at].second < v) tree[at] = pair<int, int>(id, v);
    return;
  }
  int m = (l + r) / 2;
  if (x <= H[m])
    insert(at * 2, l, m, x, v, id);
  else
    insert(at * 2 + 1, m + 1, r, x, v, id);
  if (tree[at * 2].second > tree[at * 2 + 1].second)
    tree[at] = tree[at * 2];
  else
    tree[at] = tree[at * 2 + 1];
}
pair<int, int> query(int at, int l, int r, long long int x, long long int y) {
  if (y < x) return pair<int, int>(-1, -1);
  if (x <= H[l] && H[r] <= y) return tree[at];
  if (y < H[l] || x > H[r]) return pair<int, int>(-1, -1);
  int m = (l + r) / 2;
  pair<int, int> a = query(2 * at, l, m, x, y);
  pair<int, int> b = query(2 * at + 1, m + 1, r, x, y);
  if (a.second > b.second) return a;
  return b;
}
int main() {
  int n;
  long long int d;
  for (int i = 0; i < 4 * 100005; i++) tree[i] = pair<int, int>(-1, -1);
  scanf("%d %I64d", &n, &d);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &h[i]);
    H[i] = h[i];
  }
  sort(H, H + n);
  int N = unique(H, H + n) - H;
  for (int i = 0; i < n; i++) {
    prevs[i] = -1;
    dp[i] = 1;
    pair<int, int> a = query(1, 0, N - 1, H[0], h[i] - d);
    pair<int, int> b = query(1, 0, N - 1, h[i] + d, H[N - 1]);
    if (b.second > a.second) a = b;
    if (a.first != -1) prevs[i] = a.first, dp[i] = a.second + 1;
    insert(1, 0, N - 1, h[i], dp[i], i);
  }
  int ans = 0;
  int id = 0;
  for (int i = 0; i < n; i++)
    if (dp[i] > ans) {
      ans = dp[i];
      id = i;
    }
  printf("%d\n", ans);
  vector<int> V;
  while (id != -1) {
    V.push_back(id + 1);
    id = prevs[id];
  }
  for (int i = V.size() - 1; i >= 0; i--) {
    printf("%d", V[i]);
    if (i) printf(" ");
  }
  printf("\n");
  return 0;
}
