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 maxn = 1e5 + 20;
long long mx[4 * maxn], a[maxn], pos[maxn], dp[maxn];
pair<long long, long long> h[maxn];
void update(long long p, long long value, long long s, long long e,
            long long v) {
  if (e == s) {
    mx[v] = value;
    return;
  }
  long long mid = (s + e) / 2;
  if (p <= mid)
    update(p, value, s, mid, 2 * v);
  else
    update(p, value, mid + 1, e, 2 * v + 1);
  if (mx[2 * v] > mx[2 * v + 1])
    mx[v] = mx[2 * v];
  else
    mx[v] = mx[2 * v + 1];
}
int get(long long l, long long r, long long s, long long e, long long v) {
  if (r < s || l > e) return 0;
  if (s >= l && e <= r) return mx[v];
  long long mid = (s + e) / 2;
  return max(get(l, r, s, mid, 2 * v), get(l, r, mid + 1, e, 2 * v + 1));
}
int main() {
  long long n, d;
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    h[i].first = a[i];
    h[i].second = i;
  }
  sort(h, h + n);
  for (int i = 0; i < n; i++) pos[h[i].second] = i;
  for (int i = 0; i < n; i++) update(i, -1, 0, n - 1, 1);
  long long mx = 0;
  for (int i = 0; i < n; i++) {
    pair<long long, long long> x;
    x.first = a[i] - d;
    x.second = maxn;
    int r = lower_bound(h, h + n, x) - h - 1;
    x.first = d + a[i];
    x.second = -1;
    int l = lower_bound(h, h + n, x) - h;
    int upd = max(get(0, r, 0, n - 1, 1), get(l, n - 1, 0, n - 1, 1));
    if (!i) upd = 0;
    dp[i] = upd + 1;
    update(pos[i], dp[i], 0, n - 1, 1);
    if (dp[i] > dp[mx]) mx = i;
  }
  cout << dp[mx] << endl;
  vector<int> ans;
  ans.push_back(mx);
  for (int i = mx; i >= 0; i--) {
    if (dp[i] == dp[mx] - 1 && abs(a[i] - a[mx]) >= d) {
      mx = i;
      ans.push_back(i);
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto x : ans) cout << x + 1 << " ";
}
