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 + 5, M = 3e5 + 5;
int n, d, A[N], L[N], R[N], dp[N], mx, st, sz;
long long K[N];
vector<long long> v;
pair<int, int> D[M << 2];
struct seg {
  void upd(int ind, int p, int d, int id = 1, int l = 0, int r = sz) {
    D[id] = max(D[id], {d, p});
    if (r - l < 2) return;
    int mid = (l + r) >> 1;
    if (ind < mid)
      upd(ind, p, d, id << 1, l, mid);
    else
      upd(ind, p, d, id << 1 | 1, mid, r);
  }
  pair<int, int> get(int s, int e, int id = 1, int l = 0, int r = sz) {
    if (l >= e || r <= s) return (make_pair(0, 0));
    if (l >= s && r <= e) return (D[id]);
    int mid = (l + r) >> 1;
    return (max(get(s, e, id << 1, l, mid), get(s, e, id << 1 | 1, mid, r)));
  }
} sg1;
int main() {
  scanf("%d %d", &n, &d);
  for (int i = 0; i < n; i++)
    scanf("%lld", &K[i]), v.push_back(K[i]), v.push_back(K[i] - d),
        v.push_back(K[i] + d);
  v.push_back(-1e16);
  v.push_back(1e16);
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  sz = (int)v.size();
  for (int i = 0; i < n; i++) {
    L[i] = lower_bound(v.begin(), v.end(), K[i] - d) - v.begin();
    R[i] = lower_bound(v.begin(), v.end(), K[i] + d) - v.begin();
    A[i] = lower_bound(v.begin(), v.end(), K[i]) - v.begin();
  }
  for (int i = n - 1; i >= 0; i--) {
    pair<int, int> p = {0, i};
    p = max(p, sg1.get(0, L[i] + 1));
    p = max(p, sg1.get(R[i], sz));
    p.first++;
    dp[i] = p.second;
    p.second = i;
    if (mx < p.first) mx = p.first, st = i;
    sg1.upd(A[i], p.second, p.first);
  }
  printf("%d\n", mx);
  for (int i = 0; i < mx; i++) {
    printf("%d ", st + 1);
    st = dp[st];
  }
}
