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;
map<long long, int> mp;
vector<long long> tk;
int tree[4 * 300007][3];
void update(int id, int st, int ed, int pos, pair<long long, long long> val) {
  int lft = id << 1, rgt = lft + 1, md = (st + ed) >> 1;
  if (st == pos && ed == pos) {
    if (tree[id][0] <= val.first) {
      tree[id][0] = max(tree[id][0], (int)val.first);
      tree[id][1] = val.second;
      tree[id][2] = pos;
    }
    return;
  } else {
    if (pos <= md)
      update(lft, st, md, pos, val);
    else
      update(rgt, md + 1, ed, pos, val);
    if (tree[lft][0] > tree[rgt][0]) {
      tree[id][0] = tree[lft][0];
      tree[id][1] = tree[lft][1];
      tree[id][2] = tree[lft][2];
    } else {
      tree[id][0] = tree[rgt][0];
      tree[id][1] = tree[rgt][1];
      tree[id][2] = tree[rgt][2];
    }
  }
}
pair<long long, long long> query(int id, int st, int ed, int x, int y) {
  int lft = id << 1, rgt = lft + 1, md = (st + ed) >> 1;
  if (x <= st && ed <= y) {
    return make_pair(tree[id][0], tree[id][2]);
  }
  if (x > ed || y < st)
    return make_pair(1e-15, 0);
  else if (y <= md)
    return query(lft, st, md, x, y);
  else if (x > md)
    return query(rgt, md + 1, ed, x, y);
  else
    return max(query(lft, st, md, x, y), query(rgt, md + 1, ed, x, y));
}
long long id = 1;
map<long long, long long> gmp;
int ck(int a) {
  if (a >= 1 && a <= id) return 1;
  return 0;
}
long long par[300007];
map<long long, long long> ase;
long long inp[300007];
int main() {
  long long l, u, v, w, x, y, z, a, b, c, d, e, f, t = 1, tc;
  long long flg, sz, cnt, gt, ans, mx, mn;
  long long m, k, n, i, j;
  long long low, hi, md, sm, ff, ai, bi;
  pair<long long, long long> px, py, pr;
  int p;
  memset(tree, 0, sizeof tree);
  scanf("%lld%lld", &n, &d);
  for (int i = 1; i < n + 1; i++) {
    scanf("%lld", &inp[i]);
    mp[inp[i]]++;
    mp[inp[i] + d]++;
    mp[inp[i] - d]++;
  }
  if (d == 0) {
    cout << n << endl;
    for (int i = 1; i < n + 1; i++) cout << i << " ";
    cout << endl;
    return 0;
  }
  id = 1;
  for (__typeof(mp.begin()) it = mp.begin(); it != mp.end(); it++) {
    mp[(*it).first] = id;
    id++;
  }
  z = id - 1;
  ans = -1;
  for (i = n; i >= 1; i--) {
    x = inp[i];
    ai = mp[x - d];
    bi = mp[x + d];
    p = mp[x];
    ase[p] = i;
    px = query(1, 1, z, 1, ai);
    py = query(1, 1, z, bi, z);
    if (px.first > py.first) {
      px.first++;
      update(1, 1, z, p, px);
    } else {
      py.first++;
      update(1, 1, z, p, py);
    }
    pr = max(px, py);
    if (pr.first > ans) ans = pr.first, gt = i;
    par[i] = ase[pr.second];
  }
  cout << ans << endl;
  x = gt;
  vector<long long> fnl;
  while (ans) {
    ans--;
    cout << x << " ";
    x = par[x];
  }
  return 0;
}
