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 a[100100], b[100100];
vector<long long> q;
pair<int, int> t[400400];
int p[100100];
pair<int, int> find_max(int l, int r, int lv, int rv, int v) {
  if (l > rv || r < lv) return make_pair(-1, -1);
  if (lv >= l && rv <= r) return t[v];
  return max(find_max(l, r, lv, (lv + rv) / 2, v * 2),
             find_max(l, r, (lv + rv) / 2 + 1, rv, v * 2 + 1));
}
void modify(int v, int i, int d) {
  t[v] = max(t[v], make_pair(d, i));
  v /= 2;
  for (; v > 0; v /= 2) t[v] = max(t[v * 2], t[v * 2 + 1]);
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, d;
  scanf("%d%d", &n, &d);
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; i++) b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++)
    if (i == 0 || b[i] != b[i - 1]) q.push_back(b[i]);
  int k = q.size();
  int z = 1;
  pair<int, int> ans = make_pair(-1, -1);
  while (z < k) z *= 2;
  cerr << z << endl;
  for (int i = 0; i < z * 2; i++) t[i] = make_pair(0, -1);
  for (int i = 0; i < n; i++) {
    int r = lower_bound(q.begin(), q.end(), a[i] + d) - q.begin();
    int l = upper_bound(q.begin(), q.end(), a[i] - d) - q.begin() - 1;
    pair<int, int> m = make_pair(0, -1);
    if (r < q.size()) {
      m = find_max(r + 1, z, 1, z, 1);
    }
    if (l >= 0) {
      m = max(m, find_max(1, l + 1, 1, z, 1));
    }
    int t = lower_bound(q.begin(), q.end(), a[i]) - q.begin() + 1;
    p[i] = m.second;
    modify(t + z - 1, i, m.first + 1);
    ans = max(ans, make_pair(m.first + 1, i));
  }
  printf("%d\n", ans.first);
  vector<int> A;
  while (ans.second != -1) {
    A.push_back(ans.second);
    ans.second = p[ans.second];
  }
  for (int i = A.size() - 1; i >= 0; i--) printf("%d ", A[i] + 1);
  return 0;
}
