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;
using ll = long long int;
ll hell = 10000000 + 7;
ll modexp(ll x, ll p, ll m) {
  if (p < 0) return 0;
  ll it = 1;
  while (p > 0) {
    if (p % 2 == 1) it = (it * x) % (hell);
    x = (x * x) % (hell);
    p /= 2;
  }
  return it;
}
ll modInverse(ll x, ll m) { return modexp(x, m - 2, m); }
pair<ll, ll> t[400009];
pair<ll, ll> mx(pair<ll, ll> p1, pair<ll, ll> p2) {
  if (p1.first > p2.first)
    return p1;
  else
    return p2;
}
void upd(ll i, ll s, ll e, ll pos, ll index, ll value) {
  if (s == e) {
    t[i] = mx(t[i], {value, index});
    return;
  }
  ll tm = (s + e) / 2;
  if (pos <= tm)
    upd(2 * i + 1, s, tm, pos, index, value);
  else
    upd(2 * i + 2, tm + 1, e, pos, index, value);
  t[i] = mx(t[2 * i + 1], t[2 * i + 2]);
}
pair<ll, ll> qry(ll i, ll s, ll e, ll a, ll b) {
  if (s > b || e < a) return {0, -1};
  if (s >= a && e <= b) return t[i];
  ll tm = (s + e) / 2;
  return mx(qry(2 * i + 1, s, tm, a, b), qry(2 * i + 2, tm + 1, e, a, b));
}
int main() {
  ll n, d;
  cin >> n >> d;
  ll a[n];
  for (ll i = 0; i < 40000; ++i) t[i] = {0, -1};
  set<ll> s;
  ll track[n];
  for (ll i = 0; i < n; ++i) {
    cin >> a[i];
    s.insert(a[i]);
    track[i] = -1;
  }
  map<ll, ll> m;
  ll cnt = 0;
  for (auto i : s) {
    if (m.find(i) == m.end()) {
      m[i] = cnt++;
    }
  }
  upd(0, 0, cnt, m[a[0]], 0, 1);
  ll ans = 1;
  ll tail = -1;
  for (ll i = 1; i < n; ++i) {
    auto it2 = s.lower_bound(a[i] + d);
    pair<ll, ll> mm = {0, -1};
    if (it2 != s.end()) {
      ll id2 = m[*it2];
      mm = qry(0, 0, cnt, id2, cnt);
    }
    auto it = s.upper_bound(a[i] - d);
    if (it != s.begin()) {
      it--;
      mm = mx(mm, qry(0, 0, cnt, 0, m[*it]));
    }
    if (ans < mm.first + 1) {
      ans = mm.first + 1;
      tail = i;
    }
    track[i] = mm.second;
    upd(0, 0, cnt, m[a[i]], i, mm.first + 1);
  }
  cout << ans << endl;
  vector<ll> vv;
  while (tail != -1) {
    vv.push_back(tail);
    tail = track[tail];
  }
  reverse(vv.begin(), vv.end());
  for (ll j = 0; j < vv.size(); ++j) cout << vv[j] + 1 << " ";
  return 0;
}
