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;
struct segTree {
  long long N;
  vector<long long> seg;
  segTree(long long x) : seg(4 * x), N(x) {}
  long long f(long long prev_val, long long new_val, long long assign = 0) {
    if (assign) return new_val;
    return max(prev_val, new_val);
  }
  long long Q(long long l, long long r, long long l1, long long r1,
              long long node) {
    if (l1 > r or r1 < l) return 0;
    if (l <= l1 and r1 <= r) return seg[node];
    long long mid = (l1 + r1) / 2;
    return f(Q(l, r, l1, mid, 2 * node), Q(l, r, mid + 1, r1, 2 * node + 1));
  }
  void U(long long ind, long long val, long long l1, long long r1,
         long long node) {
    if (l1 == r1) {
      if (ind != 0) seg[node] = f(seg[node], val, 1);
      return;
    }
    long long mid = (l1 + r1) / 2;
    if (ind <= mid)
      U(ind, val, l1, mid, 2 * node);
    else
      U(ind, val, mid + 1, r1, 2 * node + 1);
    seg[node] = f(seg[2 * node], seg[2 * node + 1]);
  }
  long long queri(long long l, long long r) { return Q(l, r, 1, N, 1); }
  void update(long long ind, long long val) { U(ind, val, 1, N, 1); }
};
long long n, d, h[400000], id[400000][3], dp[400000];
void compress() {
  vector<pair<long long, long long>> v;
  for (long long i = 1; i < n + 1; i++) {
    v.push_back({h[i], 3 * i});
    v.push_back({h[i] - d, 3 * i + 1});
    v.push_back({h[i] + d, 3 * i + 2});
  }
  sort(v.begin(), v.end());
  long long cur = 1;
  id[v[0].second / 3][(v[0].second) % 3] = cur;
  for (long long i = 1; i < v.size(); i++) {
    if (v[i].first != v[i - 1].first) cur++;
    id[v[i].second / 3][(v[i].second) % 3] = cur;
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> d;
  for (long long i = 1; i < n + 1; i++) cin >> h[i];
  compress();
  segTree tree(400000);
  for (long long i = 1; i < n + 1; i++) {
    dp[i] = max(tree.queri(id[i][2], 400000), tree.queri(1, id[i][1])) + 1;
    tree.update(id[i][0], dp[i]);
  }
  long long cur = 1;
  for (long long i = 2; i < n + 1; i++) {
    if (dp[i] > dp[cur]) cur = i;
  }
  vector<long long> ans;
  long long las = cur;
  ans.push_back(las);
  cur--;
  while (cur > 0) {
    if (abs(h[las] - h[cur]) >= d and dp[cur] == dp[las] - 1) {
      ans.push_back(cur);
      las = cur;
    }
    cur--;
  }
  reverse(ans.begin(), ans.end());
  cout << ans.size() << '\n';
  for (auto x : ans) cout << x << " ";
}
