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 + 3;
int n;
long long D, a[N];
int res = 0, dp[N];
pair<long long, int> b[N];
int tt[N];
long long gt[N];
vector<int> da;
struct ok {
  int lo;
  int hi;
  int val;
};
ok Node[4 * N];
void build(int id, int l, int r) {
  Node[id].lo = l;
  Node[id].hi = r;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
}
void update(int id, int pos, int val) {
  if (Node[id].lo > pos || Node[id].hi < pos) return;
  if (Node[id].lo == Node[id].hi) {
    Node[id].val = val;
    return;
  }
  update(id * 2, pos, val);
  update(id * 2 + 1, pos, val);
  Node[id].val = max(Node[id * 2].val, Node[id * 2 + 1].val);
}
int getmax(int id, int l, int r) {
  if (l > r) return 0;
  if (Node[id].lo > r || Node[id].hi < l) return 0;
  if (l <= Node[id].lo && Node[id].hi <= r) return Node[id].val;
  return max(getmax(id * 2, l, r), getmax(id * 2 + 1, l, r));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> D;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b + 1, b + n + 1);
  int dem = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i].first != b[i - 1].first) {
      dem++;
    }
    tt[b[i].second] = dem;
    gt[dem] = b[i].first;
  }
  build(1, 1, dem);
  dp[1] = 1;
  update(1, tt[1], 1);
  for (int i = 2; i <= n; i++) {
    int vt = 0;
    int lo = 1, hi = tt[i], mid;
    long long x = a[i] - D;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      if (gt[mid] <= x) {
        vt = mid;
        lo = mid + 1;
      } else
        hi = mid - 1;
    }
    dp[i] = max(dp[i], getmax(1, 1, vt) + 1);
    vt = n + 1;
    lo = tt[i];
    hi = dem;
    x = a[i] + D;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      if (gt[mid] >= x) {
        vt = mid;
        hi = mid - 1;
      } else
        lo = mid + 1;
    }
    dp[i] = max(dp[i], getmax(1, vt, dem) + 1);
    update(1, tt[i], dp[i]);
  }
  for (int i = 1; i <= n; i++) {
    res = max(res, dp[i]);
  }
  cout << res << '\n';
  long long cur = -D;
  for (int i = n; i >= 1; i--) {
    if (dp[i] == res && abs(a[i] - cur) >= D) {
      res--;
      da.push_back(i);
      cur = a[i];
    }
  }
  for (int i = da.size() - 1; i >= 0; i--) cout << da[i] << " ";
}
