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 long long int N = 1e5 + 10, N2 = 2e5 + 10, delta = 46639, mod = 1e9 + 7,
                    oo = N - 1, LOG = 20;
const long double PI = 3.141592653589793;
pair<long long int, long long int> mx[N << 2];
long long int a[N], dp[N], par[N];
vector<pair<long long int, long long int> > u;
vector<long long int> g;
void add(long long int ind, long long int val, long long int l = 0,
         long long int r = N, long long int id = 1) {
  if (r - l <= 1) {
    mx[id] = {val, ind};
    return;
  }
  long long int mid = (r + l) / 2;
  if (ind < mid)
    add(ind, val, l, mid, 2 * id);
  else
    add(ind, val, mid, r, 2 * id + 1);
  mx[id] = max(mx[2 * id], mx[2 * id + 1]);
}
pair<long long int, long long int> get(long long int l, long long int r,
                                       long long int b = 0, long long int e = N,
                                       long long int id = 1) {
  if (l <= b && e <= r) return mx[id];
  long long int mid = (b + e) / 2;
  pair<long long int, long long int> ans = {0, 0};
  if (l < mid) ans = max(ans, get(l, r, b, mid, 2 * id));
  if (mid < r) ans = max(ans, get(l, r, mid, e, 2 * id + 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n, d, maxx = 0;
  cin >> n >> d;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    maxx = max(maxx, a[i]);
    u.push_back({a[i], i});
    par[i] = -1;
  }
  sort(u.begin(), u.end());
  for (long long int i = 0; i < n; i++) {
    dp[i] = 1;
    long long int ind =
        lower_bound(u.begin(), u.end(), make_pair(a[i], i)) - u.begin();
    if (!i) {
      add(ind, dp[i]);
      continue;
    }
    long long int x =
        upper_bound(u.begin(), u.end(), make_pair(a[i] - d, n)) - u.begin();
    if (x) {
      pair<long long int, long long int> ans = get(0, x);
      par[i] = u[ans.second].second;
      dp[i] = ans.first + 1;
    }
    if (a[i] + d <= maxx) {
      long long int y =
          lower_bound(u.begin(), u.end(), make_pair(a[i] + d, 0ll)) - u.begin();
      pair<long long int, long long int> ans = get(y, n);
      dp[i] = max(dp[i], ans.first + 1);
      if (dp[i] == ans.first + 1) par[i] = u[ans.second].second;
    }
    add(ind, dp[i]);
  }
  long long int ans = 0, x = 0;
  for (long long int i = 0; i < n; i++) {
    ans = max(ans, dp[i]);
    if (ans == dp[i]) x = i;
  }
  cout << ans << '\n';
  g.push_back(x);
  while (dp[x] != 1) {
    x = par[x];
    g.push_back(x);
  }
  reverse(g.begin(), g.end());
  for (auto u : g) cout << u + 1 << ' ';
  cout << '\n';
  return 0;
}
