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 = 1e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
long long h[N], dp[N], par[N], n;
pair<long long, long long> a[N], seg[4 * N];
void build(int l = 0, int r = n, int id = 1) {
  if (r - l == 1) {
    seg[id] = make_pair(0LL, a[l].second);
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * id);
  build(mid, r, 2 * id + 1);
  seg[id] = max(seg[2 * id], seg[2 * id + 1]);
}
void upd(int x, int val, int l = 0, int r = n, int id = 1) {
  if (r - l == 1) {
    seg[id].first = 1LL * val;
    return;
  }
  int mid = (l + r) / 2;
  if (x < mid)
    upd(x, val, l, mid, 2 * id);
  else
    upd(x, val, mid, r, 2 * id + 1);
  seg[id] = max(seg[2 * id], seg[2 * id + 1]);
}
pair<long long, long long> get(int l, int r, int s = 0, int e = n, int id = 1) {
  if (e <= l || r <= s) return make_pair(0LL, 0LL);
  if (l <= s && e <= r) return seg[id];
  int mid = (s + e) / 2;
  return max(get(l, r, s, mid, 2 * id), get(l, r, mid, e, 2 * id + 1));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d;
  cin >> n >> d;
  for (int i = 0; i < n; i++) cin >> h[i], a[i] = make_pair(h[i], 1LL * i);
  sort(a, a + n);
  build();
  for (int i = 0; i < n; i++) {
    int x = lower_bound(a, a + n, make_pair(h[i] + d, -inf)) - a;
    int y = lower_bound(a, a + n, make_pair(h[i] - d, inf)) - a;
    int z = lower_bound(a, a + n, make_pair(h[i], 1LL * i)) - a;
    pair<long long, long long> res = max(get(0, y), get(x, n));
    dp[i] = res.first + 1;
    upd(z, dp[i]);
    par[i] = (dp[i] == 1 ? -1 : res.second);
  }
  int mx = 0;
  for (int i = 0; i < n; i++)
    if (dp[mx] < dp[i]) mx = i;
  vector<int> ans;
  while (mx != -1) {
    ans.push_back(mx);
    mx = par[mx];
  }
  reverse((ans).begin(), (ans).end());
  cout << (int)(ans.size()) << '\n';
  for (int x : ans) cout << x + 1 << ' ';
  return 0;
}
