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>
const int M = 4e5 + 10;
using namespace std;
long long th[M], h[M];
int dp[M], st[M], par[M];
int cnt;
pair<int, int> seg[M];
long long *it;
pair<int, int> query(int id, int b, int e, int l, int r) {
  if (l == b && r == e) return seg[id];
  int mid = (b + e) >> 1;
  if (r <= mid) return query(id * 2, b, mid, l, r);
  if (l > mid) return query(id * 2 + 1, mid + 1, e, l, r);
  return max(query(id * 2, b, mid, l, mid),
             query(id * 2 + 1, mid + 1, e, mid + 1, r));
}
void update(int id, int b, int e, int pos, int val, int ind) {
  if (b == e) {
    seg[id] = max(seg[id], {val, ind});
    return;
  }
  int mid = (b + e) >> 1;
  if (pos <= mid)
    update(id * 2, b, mid, pos, val, ind);
  else
    update(id * 2 + 1, mid + 1, e, pos, val, ind);
  seg[id] = max(seg[id * 2], seg[id * 2 + 1]);
}
pair<int, int> left(long long x) {
  if (x < th[1]) return {0, 0};
  return query(1, 1, cnt, 1, upper_bound(th + 1, it, x) - 1 - th);
}
pair<int, int> right(long long x) {
  if (x > *(it - 1)) return {0, 0};
  return query(1, 1, cnt, lower_bound(th + 1, it, x) - th, cnt);
}
void prep(int n) {
  sort(th + 1, th + n + 1);
  it = unique(th + 1, th + n + 1);
  cnt = it - th - 1;
  for (int i = 1; i <= n; i++) {
    st[i] = lower_bound(th + 1, it, h[i]) - th;
  }
}
void get_dp(int n, int d) {
  for (int i = 1; i <= n; i++) {
    pair<int, int> temp = max(left(h[i] - d), right(h[i] + d));
    dp[i] = temp.first + 1;
    par[i] = temp.second;
    update(1, 1, cnt, st[i], dp[i], i);
  }
}
int main() {
  int n, d;
  cin >> n >> d;
  for (int i = 1; i <= n; i++) cin >> h[i], th[i] = h[i];
  prep(n);
  get_dp(n, d);
  int ans = 0, curr = 0;
  for (int i = 1; i <= n; i++)
    if (dp[i] > ans) ans = dp[i], curr = i;
  vector<int> vec;
  while (ans > 0) vec.push_back(curr), curr = par[curr], ans--;
  reverse(vec.begin(), vec.end());
  cout << vec.size() << endl;
  for (int i : vec) cout << i << " ";
}
