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;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
const long long maxn = 1e5 + 10;
vector<long long> comp;
pair<long long, long long> seg[4 * maxn];
pair<long long, long long> dp[maxn];
map<long long, long long> m;
void add(long long id, long long l, long long r, long long idx,
         pair<long long, long long> val) {
  if (idx < l || idx >= r) {
    return;
  }
  if (r == l + 1) {
    seg[id] = max(seg[id], val);
    return;
  }
  long long mid = (l + r) / 2;
  add(2 * id, l, mid, idx, val);
  add(2 * id + 1, mid, r, idx, val);
  seg[id] = max(seg[2 * id], seg[2 * id + 1]);
}
pair<long long, long long> get(long long id, long long l, long long r,
                               long long L, long long R) {
  if (L >= R || (R <= l || r <= L)) {
    return {0, 0};
  }
  if (l >= L && r <= R) {
    return seg[id];
  }
  long long mid = (l + r) / 2;
  return max(get(2 * id, l, mid, L, R), get(2 * id + 1, mid, r, L, R));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, d;
  cin >> n >> d;
  long long a[n];
  long long h[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    h[i] = a[i];
    comp.push_back(a[i]);
  }
  sort(comp.begin(), comp.end());
  for (long long i = 0; i < comp.size(); i++) {
    m[comp[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    a[i] = m[a[i]];
  }
  for (long long i = n - 1; i >= 0; i--) {
    long long x =
        lower_bound(comp.begin(), comp.end(), h[i] + d) - comp.begin();
    dp[i] = get(1, 0, n, x, n);
    x = upper_bound(comp.begin(), comp.end(), h[i] - d) - comp.begin();
    dp[i] = max(dp[i], get(1, 0, n, 0, x));
    dp[i].first++;
    add(1, 0, n, a[i], make_pair(dp[i].first, i));
  }
  long long maxi = dp[0].first;
  long long boz = 0;
  for (long long i = 1; i < n; i++) {
    if (dp[i].first > maxi) {
      maxi = dp[i].first;
      boz = i;
    }
  }
  vector<long long> ans;
  for (long long i = 0; i < maxi; i++) {
    ans.push_back(boz);
    boz = dp[boz].second;
  }
  cout << maxi << endl;
  for (long long i : ans) {
    cout << i + 1 << ' ';
  }
}
