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 M = 1e5 + 5;
const int mod = 1e9 + 7;
const double esp = 1e-10;
const long double pi = 4 * atan((long double)1);
int n;
long long d;
struct mido {
  long long l, r, val;
} a[M];
vector<pair<long long, int> > v;
pair<int, int> f[12 * M];
int trace[M];
set<long long> s;
void upd(int id, int l, int r, int i, int k, int vt) {
  if (l > r) return;
  if (r < i || i < l) return;
  if (l == r) {
    if (f[id].first < k) f[id] = pair<int, int>(k, vt);
    return;
  }
  int mid = (l + r) / 2;
  upd(id * 2, l, mid, i, k, vt);
  upd(id * 2 + 1, mid + 1, r, i, k, vt);
  if (f[id * 2].first > f[id * 2 + 1].first) {
    f[id] = f[id * 2];
  } else
    f[id] = f[id * 2 + 1];
}
pair<int, int> get(int id, int l, int r, int u, int v) {
  if (l > r) return pair<int, int>(0, 0);
  if (r < u || v < l) return pair<int, int>(0, 0);
  if (u <= l && r <= v) {
    return f[id];
  }
  int mid = (l + r) / 2;
  pair<int, int> k1 = get(id * 2, l, mid, u, v);
  pair<int, int> k2 = get(id * 2 + 1, mid + 1, r, u, v);
  if (k1.first > k2.first) return k1;
  return k2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> d;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].val;
    v.push_back(pair<long long, int>(a[i].val, i));
    v.push_back(pair<long long, int>(a[i].val - d, i + n));
    v.push_back(pair<long long, int>(a[i].val + d, i + 2 * n));
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < 3 * n; i++) {
    s.insert(v[i].first);
    int vt = v[i].second;
    if (vt <= n)
      a[vt].val = s.size();
    else if (vt <= 2 * n)
      a[vt - n].l = s.size();
    else
      a[vt - 2 * n].r = s.size();
  }
  int ans = 0, last;
  for (int i = 1; i <= n; i++) {
    pair<int, int> k1 = get(1, 1, 3 * n, 1, a[i].l);
    pair<int, int> k2 = get(1, 1, 3 * n, a[i].r, 3 * n);
    if (k1.first > k2.first) {
      upd(1, 1, 3 * n, a[i].val, k1.first + 1, i);
      trace[i] = k1.second;
      if (ans < k1.first + 1) {
        ans = k1.first + 1;
        last = i;
      }
    } else {
      upd(1, 1, 3 * n, a[i].val, k2.first + 1, i);
      trace[i] = k2.second;
      if (ans < k2.first + 1) {
        ans = k2.first + 1;
        last = i;
      }
    }
  }
  cout << ans << endl;
  vector<int> res;
  while (last > 0) {
    res.push_back(last);
    last = trace[last];
  }
  for (int i = res.size() - 1; i >= 0; i--) cout << res[i] << " ";
  return 0;
}
