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 N = 131072;
int tab[2 * N];
int previ[N];
void insert(int x, int v) {
  x += N;
  tab[x] = max(tab[x], v);
  x /= 2;
  while (x) {
    tab[x] = max(tab[2 * x], tab[2 * x + 1]);
    x /= 2;
  }
}
int query(int a, int b) {
  a += N, b += N;
  int ans = max(tab[a], tab[b]);
  while (a / 2 != b / 2) {
    if (a % 2 == 0) ans = max(ans, tab[a + 1]);
    if (b % 2 == 1) ans = max(ans, tab[b - 1]);
    a /= 2;
    b /= 2;
  }
  return ans;
}
int go_down(int x, int m) {
  while (x < N) {
    if (tab[x * 2] == m)
      x = x * 2;
    else
      x = x * 2 + 1;
  }
  return x - N;
}
int find_max(int a, int b, int m) {
  a += N, b += N;
  if (tab[a] == m) return go_down(a, m);
  if (tab[b] == m) return go_down(b, m);
  while (a / 2 != b / 2) {
    if (a % 2 == 0 && tab[a + 1] == m) return go_down(a + 1, m);
    if (b % 2 == 1 && tab[b - 1] == m) return go_down(b - 1, m);
    a /= 2;
    b /= 2;
  }
  assert(0);
  return 3;
}
void update(int &ans, int i, int a, int b) {
  int m = query(a, b);
  if (m + 1 > ans) {
    ans = m + 1;
    int pre = find_max(a, b, m);
    previ[i] = pre;
  }
}
int main() {
  int n, d;
  scanf("%d %d", &n, &d);
  set<long long> S;
  vector<long long> _W(n);
  for (int i = 0; i < (n); ++i) {
    long long h;
    scanf("%I64d", &h);
    S.insert(h);
    _W[i] = h;
  }
  vector<long long> RM(S.begin(), S.end());
  map<long long, int> M;
  for (int i = 0; i < ((int)RM.size()); ++i) M[RM[i]] = i;
  vector<int> W(n);
  for (int i = 0; i < (n); ++i) {
    W[i] = M[_W[i]];
  }
  int best = 0;
  int last = -1;
  for (int i = 0; i < (n); ++i) {
    auto &&it = S.upper_bound(_W[i] - d);
    int ans = 0;
    if (it != S.begin()) {
      --it;
      int h = M[*it];
      update(ans, i, 0, h);
    }
    auto &&it2 = S.lower_bound(_W[i] + d);
    if (it2 != S.end()) {
      int h = M[*it2];
      update(ans, i, h, N - 1);
    }
    if (best < ans) {
      best = ans;
      last = i;
    }
    insert(W[i], ans);
  }
  printf("%d\n", best);
  vector<int> A(best);
  while (best) {
    A[best - 1] = last;
    best--;
    if (best) {
      long long tr = RM[previ[last]];
      int i = last - 1;
      while (_W[i] != tr) i--;
      last = i;
    }
  }
  for (auto &&l : A) printf("%d ", l + 1);
  printf("\n");
  return 0;
}
