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;
void file() {}
const int N = 1e6 + 5, MAX = 3e7, MOD = 1e9 + 7;
int n, d;
long long a[N];
int pos[N], dp[N], st = -1, camefrom[N], aftersort[N];
pair<int, int> seg[4 * N];
void update(int indx, int l, int r, int pos, int val) {
  if (l == r) {
    seg[indx] = {val, l};
    return;
  }
  if (pos <= (l + r) / 2)
    update(indx * 2, l, (l + r) / 2, pos, val);
  else
    update(indx * 2 + 1, (l + r) / 2 + 1, r, pos, val);
  seg[indx] = max(seg[indx * 2], seg[indx * 2 + 1]);
}
pair<int, int> get(int indx, int l, int r, int i, int j) {
  if (l > j || r < i) return {0, 0};
  if (l >= i && r <= j) return seg[indx];
  return max(get(indx * 2, l, (l + r) / 2, i, j),
             get(indx * 2 + 1, (l + r) / 2 + 1, r, i, j));
}
int mx;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  file();
  scanf("%d %d", &n, &d);
  vector<pair<long long, int>> b(n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]), b[i] = {a[i], i};
  sort((b).begin(), (b).end());
  memset(camefrom, -1, sizeof(camefrom));
  for (int i = 0; i < n; i++) pos[b[i].second] = i, aftersort[i] = b[i].second;
  for (int i = n - 1; i >= 0; i--) {
    auto it =
        upper_bound((b).begin(), (b).end(), make_pair(a[i] - d, 1000000000));
    dp[pos[i]] = 1;
    if (st == -1) st = i;
    if (it != b.begin()) {
      it--;
      int poss = it - b.begin();
      pair<int, int> p = get(1, 0, n - 1, 0, poss);
      if (p.first + 1 > dp[pos[i]]) {
        dp[pos[i]] = p.first + 1;
        camefrom[i] = aftersort[p.second];
        if (dp[pos[i]] > mx) {
          mx = dp[pos[i]];
          st = i;
        }
      }
    }
    it = lower_bound((b).begin(), (b).end(), make_pair(d + a[i], 0));
    if (it != b.end()) {
      int poss = it - b.begin();
      pair<int, int> p = get(1, 0, n - 1, poss, n - 1);
      if (p.first + 1 > dp[pos[i]]) {
        dp[pos[i]] = p.first + 1;
        camefrom[i] = aftersort[p.second];
        if (dp[pos[i]] > mx) {
          mx = dp[pos[i]];
          st = i;
        }
      }
    }
    update(1, 0, n - 1, pos[i], dp[pos[i]]);
  }
  printf("%d\n", dp[pos[st]]);
  int curr = st;
  while (curr != -1) {
    printf("%d ", curr + 1);
    curr = camefrom[curr];
  }
}
