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 long long mod = 1e9 + 7;
const long long maxn = 1e5 + 5;
pair<long long, long long> t[4 * maxn];
void update(long long v, long long l, long long r, long long x,
            pair<long long, long long> val) {
  if (l > r) return;
  if (l > x or r < x) return;
  if (l == x and r == x) {
    if (t[v].first > val.first) return;
    t[v] = val;
    return;
  }
  update(2 * v, l, (l + r) / 2, x, val);
  update(2 * v + 1, (l + r) / 2 + 1, r, x, val);
  if (t[v].first < t[2 * v].first) t[v] = t[2 * v];
  if (t[v].first < t[2 * v + 1].first) t[v] = t[2 * v + 1];
}
pair<long long, long long> find(long long v, long long l, long long r,
                                long long x, long long y) {
  if (l > r) return {0, 0};
  if (l > y or r < x) return {0, 0};
  if (x <= l and y >= r) return t[v];
  auto i = find(2 * v, l, (l + r) / 2, x, y);
  auto j = find(2 * v + 1, (l + r) / 2 + 1, r, x, y);
  if (i.first > j.first)
    return i;
  else
    return j;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, d;
  cin >> n >> d;
  long long h[n], nextid[n];
  set<long long> s;
  for (long long i = 0; i < n; ++i) {
    cin >> h[i];
    s.insert(h[i]);
  }
  map<long long, long long> m;
  long long j = 0;
  for (auto &it : s) {
    m[it] = j;
    j++;
  }
  j--;
  long long ans = 0, startindex = n;
  for (long long i = n - 1; i >= 0; i--) {
    auto it = s.lower_bound(h[i] + d);
    long long val = 1;
    nextid[i] = -1;
    if (it != s.end()) {
      auto temp = find(1, 0, j, m[*it], j);
      if (val < 1 + temp.first) {
        val = 1 + temp.first;
        nextid[i] = temp.second;
      }
    }
    it = s.upper_bound(h[i] - d);
    if (it != s.begin()) {
      it--;
      auto temp = find(1, 0, j, 0, m[*it]);
      if (val < 1 + temp.first) {
        val = 1 + temp.first;
        nextid[i] = temp.second;
      }
    }
    update(1, 0, j, m[h[i]], {val, i});
    if (ans < val) {
      ans = val;
      startindex = i;
    }
    ans = max(ans, val);
  }
  cout << ans << '\n';
  ;
  while (startindex != -1) {
    cout << startindex + 1 << ' ';
    startindex = nextid[startindex];
  }
  cout << '\n';
  ;
  return 0;
}
