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 = 100005;
int qs, qe, q;
int val;
int ST[4 * N + 10];
void update(int x, int y, int n) {
  if (q > y || q < x)
    return;
  else if (x == y && q == x)
    ST[n] = max(ST[n], val);
  else {
    update(x, (x + y) / 2, 2 * n);
    update((x + y) / 2 + 1, y, 2 * n + 1);
    ST[n] = max(ST[2 * n], ST[2 * n + 1]);
  }
}
int query(int x, int y, int n) {
  if (qs > y || qe < x)
    return 0;
  else if (x >= qs && y <= qe)
    return ST[n];
  else
    return max(query(x, (x + y) / 2, 2 * n),
               query((x + y) / 2 + 1, y, 2 * n + 1));
}
int main() {
  int n;
  long long d;
  cin >> n >> d;
  vector<long long> h(n);
  for (int i = 0; i < n; i++) cin >> h[i];
  vector<long long> sh = h;
  sort(sh.begin(), sh.end());
  vector<long long> rh = sh;
  reverse(rh.begin(), rh.end());
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    if (i == 0)
      mp[sh[i]] = 1;
    else
      mp[sh[i]] = (sh[i] == sh[i - 1] ? mp[sh[i - 1]] : mp[sh[i - 1]] + 1);
  }
  vector<int> dp(n);
  dp[0] = 1;
  q = mp[h[0]];
  val = 1;
  update(1, N, 1);
  for (int i = 1; i < n; i++) {
    long long s = h[i] - d;
    int lo = 0, hi = n;
    while (lo < hi) {
      int m = (lo + hi) / 2;
      if (s < sh[m])
        hi = m;
      else
        lo = m + 1;
    }
    if (lo) {
      qs = 1;
      qe = mp[sh[lo - 1]];
      dp[i] = max(dp[i], query(1, N, 1) + 1);
    }
    s = h[i] + d;
    lo = 0, hi = n;
    while (lo < hi) {
      int m = (lo + hi) / 2;
      if (s > rh[m])
        hi = m;
      else
        lo = m + 1;
    }
    if (lo) {
      qs = mp[rh[lo - 1]];
      qe = N;
      dp[i] = max(dp[i], query(1, N, 1) + 1);
    }
    q = mp[h[i]];
    val = dp[i];
    update(1, N, 1);
  }
  int bst = 0;
  for (int i = 1; i < n; i++)
    if (dp[i] > dp[bst]) bst = i;
  int cur = bst;
  vector<int> ans;
  ans.push_back(cur);
  for (int i = bst - 1; i >= 0; i--) {
    if (abs(h[i] - h[cur]) >= d && dp[i] + 1 == dp[cur]) {
      ans.push_back(i);
      cur = i;
    }
  }
  cout << ans.size() << endl;
  for (int i = (int)ans.size() - 1; i >= 0; i--) cout << (ans[i] + 1) << ' ';
  cout << endl;
}
