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>
template <typename T>
T in() {
  char ch;
  T n = 0;
  bool ng = false;
  while (1) {
    ch = getchar();
    if (ch == '-') {
      ng = true;
      ch = getchar();
      break;
    }
    if (ch >= '0' && ch <= '9') break;
  }
  while (1) {
    n = n * 10 + (ch - '0');
    ch = getchar();
    if (ch < '0' || ch > '9') break;
  }
  return (ng ? -n : n);
}
using namespace std;
long long int ar[1300007], d, n, mxval;
vector<long long int> v;
map<long long int, int> mp;
struct z {
  long long int val, poss;
};
z tree[1300007];
void update(int id, int l, int r, int pos, z tp) {
  if (l == r) {
    if (tree[id].val < tp.val) {
      tree[id] = tp;
    }
    return;
  }
  int mid = (l + r) / 2;
  int lft = 2 * id;
  int rgt = lft + 1;
  if (pos <= mid) {
    update(lft, l, mid, pos, tp);
  } else
    update(rgt, mid + 1, r, pos, tp);
  if (tree[lft].val > tree[rgt].val)
    tree[id] = tree[lft];
  else
    tree[id] = tree[rgt];
  return;
}
z query(int id, int l, int r, int st, int ed) {
  if (l == st && r == ed) return tree[id];
  int mid = (l + r) / 2;
  int lft = 2 * id;
  int rgt = lft + 1;
  if (ed <= mid) {
    return query(lft, l, mid, st, ed);
  } else if (st > mid) {
    return query(rgt, mid + 1, r, st, ed);
  } else {
    z a, b, c;
    a = query(lft, l, mid, st, mid);
    b = query(rgt, mid + 1, r, mid + 1, ed);
    if (a.val > b.val)
      c = a;
    else
      c = b;
    return c;
  }
}
int parent[1300007], len[1300007];
vector<int> anss;
map<long long int, int> mpp;
int main() {
  cin >> n >> d;
  for (int i = 1; i < n + 1; i++) {
    cin >> ar[i];
    mp[ar[i]] = mp[ar[i] - d] = mp[ar[i] + d] = 1;
  }
  int cnt = 0, ans = 0, st = 0;
  for (map<long long int, int>::iterator it = mp.begin(); it != mp.end();
       it++) {
    it->second = ++cnt;
  }
  mxval = cnt;
  for (int i = 1; i < mxval + 1; i++) {
    z tpp;
    tpp.poss = i, tpp.val = 0;
    update(1, 1, mxval, i, tpp);
  }
  int last = 0;
  for (int i = 1; i < n + 1; i++) {
    parent[i] = i;
    len[i] = 1;
    z a, b, c, tp;
    a = query(1, 1, mxval, 1, mp[ar[i] - d]);
    b = query(1, 1, mxval, mp[ar[i] + d], mxval);
    if (a.val > b.val)
      c = a;
    else
      c = b;
    if (c.val + 1 > 1) {
      len[i] = c.val + 1;
      parent[i] = c.poss;
    }
    ans = max(ans, len[i]);
    if (ans <= len[i]) {
      ans = len[i];
      last = i;
    }
    tp.val = c.val + 1, tp.poss = i;
    update(1, 1, mxval, mp[ar[i]], tp);
  }
  cout << ans << endl;
  while (1) {
    anss.push_back(last);
    if (parent[last] == last) break;
    last = parent[last];
  }
  for (int i = anss.size() - 1; i >= 0; i--) {
    cout << anss[i] << " ";
  }
  return 0;
}
