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;
const int N = 1e5 + 100;
pair<int, int> it[4 * 3 * N];
int Trace[N];
long long a[N];
void update(int x, int i, int j, int pos, int val, int vt) {
  if (i > pos || j < pos) return;
  if (i == j && j == pos) {
    it[x] = {val, vt};
    return;
  }
  int mid = (i + j) / 2;
  update(2 * x, i, mid, pos, val, vt);
  update(2 * x + 1, mid + 1, j, pos, val, vt);
  it[x] = max(it[2 * x], it[2 * x + 1]);
}
pair<int, int> max(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first) return a;
  return b;
}
pair<int, int> get(int x, int i, int j, int l, int r) {
  if (i > r || j < l) return pair<int, int>(0, 0);
  if (l <= i && j <= r) return it[x];
  int mid = (i + j) / 2;
  return max(get(2 * x, i, mid, l, r), get(2 * x + 1, mid + 1, j, l, r));
}
int n, d, cnt;
map<long long, int> M;
set<long long> st;
void tv(int u) {
  if (u == 0) return;
  tv(Trace[u]);
  cout << u << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> d;
  for (int i = 1; i <= n; ++i)
    cin >> a[i], st.insert(a[i]), st.insert(a[i] - d), st.insert(a[i] + d);
  int cnt = 0;
  while (!st.empty()) {
    M[*st.begin()] = ++cnt;
    st.erase(st.begin());
  }
  update(1, 1, cnt, M[a[1]], 1, 1);
  pair<int, int> ans = pair<int, int>(1, 1), tmp;
  for (int i = 2; i <= n; ++i) {
    tmp = max(get(1, 1, cnt, 1, M[a[i] - d]), get(1, 1, cnt, M[a[i] + d], cnt));
    if (ans.first < tmp.first + 1) {
      ans = {tmp.first + 1, i};
    }
    update(1, 1, cnt, M[a[i]], tmp.first + 1, i);
    Trace[i] = tmp.second;
  }
  cout << ans.first << '\n';
  tv(ans.second);
  return 0;
}
