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;
long long n, k, a[200005], ls[2000050], ans[200005], l, r, p, pown = 1, anss,
                                                              indx;
map<long long, long long> mp;
vector<long long> v, an;
pair<long long, long long> t[4000006];
void upd(long long x) {
  if (!x) return;
  t[x] = max(t[2 * x], t[2 * x + 1]);
  upd(x / 2);
}
pair<long long, long long> cnt(long long x, long long L, long long R) {
  if (L > r || R < l) return make_pair(0, 0);
  if (L >= l && R <= r) return t[x];
  pair<long long, long long> k1 = cnt(2 * x, L, (L + R) / 2);
  pair<long long, long long> k2 = cnt(2 * x + 1, (L + R) / 2 + 1, R);
  return max(k1, k2);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  while (pown <= 3 * n) pown *= 2;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
    v.push_back(a[i] - k);
    v.push_back(a[i] + k);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    if (!i || v[i] != v[i - 1]) p++;
    mp[v[i]] = p;
  }
  for (int i = 1; i <= n; i++) {
    l = 1;
    r = mp[a[i] - k];
    pair<long long, long long> p1 = cnt(1, 1, pown);
    l = mp[a[i] + k];
    r = 3 * n;
    pair<long long, long long> p2 = cnt(1, 1, pown);
    p1 = max(p1, p2);
    ans[i] = p1.first + 1;
    ls[i] = p1.second;
    t[pown + mp[a[i]] - 1] = make_pair(ans[i], (long long)i);
    upd((pown + mp[a[i]] - 1) / 2);
    if (ans[i] > anss) {
      anss = ans[i];
      indx = i;
    }
  }
  while (indx) {
    an.push_back(indx);
    indx = ls[indx];
  }
  cout << an.size() << '\n';
  reverse(an.begin(), an.end());
  for (int i = 0; i < an.size(); i++) {
    cout << an[i] << " ";
  }
  return 0;
}
