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 int pot = 131072;
long long tab[100001];
int skok[100001];
int skad[100001];
int drzewko[2 * pot];
int drzewko2[2 * pot];
vector<long long> pil;
vector<int> wyn;
void wrzuc(int gdzie, int w, int p) {
  gdzie += pot - 1;
  if (drzewko[gdzie] < w) {
    while (gdzie > 0) {
      if (w > drzewko[gdzie]) {
        drzewko[gdzie] = w;
        drzewko2[gdzie] = p;
      }
      gdzie /= 2;
    }
  }
}
pair<int, int> ile(int gdzie, int pp, int kp, int pz, int kz) {
  if (pz <= pp && kz >= kp) return make_pair(drzewko[gdzie], drzewko2[gdzie]);
  int r = (pp + kp) / 2;
  pair<int, int> x = make_pair(0, 0);
  if (pz <= r) x = max(x, ile(2 * gdzie, pp, r, pz, kz));
  if (kz > r) x = max(x, ile(2 * gdzie + 1, r + 1, kp, pz, kz));
  return x;
}
int main() {
  map<long long, int> cz;
  int naj = 0, poz = 0;
  int n;
  long long d;
  scanf("%d%lld", &n, &d);
  for (int i = 1; i <= n; i++) {
    long long a;
    scanf("%lld", &a);
    tab[i] = a;
    if (cz[a] == 0) {
      pil.push_back(a);
      cz[a] = 1;
    }
  }
  pil.push_back(0);
  sort(pil.begin(), pil.end());
  for (int i = 1; i <= n; i++) {
    long long k = tab[i] + d;
    int pocz = 0, kon = pil.size();
    while (pocz < kon) {
      int s = (pocz + kon) / 2;
      if (pil[s] < k)
        pocz = s + 1;
      else
        kon = s;
    }
    pair<int, int> x = ile(1, 1, pot, pocz, pot);
    k = tab[i] - d;
    pocz = 0, kon = pil.size();
    while (pocz < kon) {
      int s = (pocz + kon + 1) / 2;
      if (pil[s] > k)
        kon = s - 1;
      else
        pocz = s;
    }
    if (pocz >= 1) x = max(x, ile(1, 1, pot, 1, pocz));
    skok[i] = x.first + 1;
    skad[i] = x.second;
    if (skok[i] > naj) {
      naj = skok[i];
      poz = i;
    }
    k = tab[i];
    pocz = 0, kon = pil.size() - 1;
    while (pocz < kon) {
      int s = (pocz + kon + 1) / 2;
      if (pil[s] > k)
        kon = s - 1;
      else
        pocz = s;
    }
    wrzuc(pocz, skok[i], i);
  }
  while (poz != 0) {
    wyn.push_back(poz);
    poz = skad[poz];
  }
  printf("%d\n", naj);
  for (int i = naj - 1; i >= 0; i--) printf("%d ", wyn[i]);
}
