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;
long long fro[400001];
long long z, tmp;
long long tr[400001], tr2[400001];
long long n, d;
vector<long long> vec;
void upd(long long t, long long l, long long r, long long L, long long R,
         long long x) {
  if (l > R || L > r || L > R) return;
  if (L <= l && r <= R) {
    if (tr[t] < x) {
      tr[t] = x;
      tr2[t] = z;
    }
    return;
  }
  if (tr[t] > tr[2 * t]) {
    tr[2 * t] = max(tr[2 * t], tr[t]);
    tr2[2 * t] = tr2[t];
  }
  if (tr[t] > tr[2 * t + 1]) {
    tr[2 * t + 1] = max(tr[2 * t + 1], tr[t]);
    tr2[2 * t + 1] = tr2[t];
  }
  upd(2 * t, l, (l + r) / 2, L, R, x);
  upd(2 * t + 1, (l + r) / 2 + 1, r, L, R, x);
}
long long get_max(long long t, long long l, long long r, long long L,
                  long long R) {
  if (l > R || L > r) return 0;
  if (l == r) fro[l] = tr2[t];
  if (L <= l && r <= R) return tr[t];
  if (tr[t] > tr[2 * t]) {
    tr[2 * t] = max(tr[2 * t], tr[t]);
    tr2[2 * t] = tr2[t];
  }
  if (tr[t] > tr[2 * t + 1]) {
    tr[2 * t + 1] = max(tr[2 * t + 1], tr[t]);
    tr2[2 * t + 1] = tr2[t];
  }
  return max(get_max(2 * t, l, (l + r) / 2, L, R),
             get_max(2 * t + 1, (l + r) / 2 + 1, r, L, R));
}
pair<long long, long long> P[400001];
long long indx[400001];
long long i, ans;
int main() {
  cin >> n >> d;
  for (i = 1; i <= n; i++) {
    cin >> P[i].first;
    P[i].second = i;
  }
  z = 0;
  upd(1, 1, n, 1, n, 1);
  sort(P + 1, P + n + 1);
  for (i = 1; i <= n; i++) indx[P[i].second] = i;
  for (long long j = 1; j <= n; j++) {
    i = indx[j];
    z = i;
    long long l = 1, r = i - 1, m = 0, lst = 0;
    long long now = get_max(1, 1, n, i, i);
    if (now > ans) {
      ans = now;
      tmp = i;
    }
    while (l <= r) {
      m = (l + r) / 2;
      if (P[i].first - P[m].first >= d) {
        lst = m;
        l = m + 1;
      } else
        r = m - 1;
    }
    if (lst != 0) upd(1, 1, n, 1, lst, now + 1);
    l = i;
    r = n;
    lst = n + 1;
    while (l <= r) {
      m = (l + r) / 2;
      if (P[m].first - P[i].first >= d) {
        lst = m;
        r = m - 1;
      } else
        l = m + 1;
    }
    upd(1, 1, n, lst, n, now + 1);
  }
  while (tmp != 0) {
    vec.push_back(tmp);
    tmp = fro[tmp];
  }
  reverse(vec.begin(), vec.end());
  cout << vec.size() << endl;
  for (i = 0; i < vec.size(); i++) cout << P[vec[i]].second << " ";
  cout << endl;
}
