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 = 6e6 + 5;
long long a[N], BIG = 1e15 + 1;
int LL[N], RR[N], ID = 1, ans[N], par[N], val, indx;
pair<int, int> seg[N];
void up(long long p, long long L, long long R, int id) {
  int lid = LL[id], rid = RR[id];
  if (L == R) {
    seg[id] = {val, indx};
    return;
  }
  long long m = (L + R) >> 1;
  if (p <= m && p >= L) {
    if (lid == 0) {
      lid = ++ID;
      LL[id] = lid;
    }
    up(p, L, m, lid);
  } else if (p <= R && p > m) {
    if (rid == 0) {
      rid = ++ID;
      RR[id] = rid;
    }
    up(p, m + 1, R, rid);
  }
  seg[id] = max(seg[lid], seg[rid]);
}
pair<int, int> get(long long l, long long r, long long L, long long R, int id) {
  if (R < l || r < L || id == 0) return {0, -1};
  int lid = LL[id], rid = RR[id];
  if (l <= L && R <= r) return seg[id];
  long long m = (L + R) >> 1ll;
  return max(get(l, r, L, m, lid), get(l, r, m + 1, R, rid));
}
int main() {
  int n, d;
  cin >> n >> d;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  int mx = -1, idx;
  for (int i = 1; i <= n; i++) {
    pair<int, int> x = {0, -1};
    if (a[i] - d > 0) x = get(1, a[i] - d, 1, BIG, 1);
    if (a[i] + d < BIG) x = max(x, get(a[i] + d, BIG, 1, BIG, 1));
    par[i] = x.second;
    ans[i] = x.first + 1;
    if (mx < ans[i]) {
      mx = ans[i], idx = i;
    }
    val = ans[i], indx = i;
    up(a[i], 1, BIG, 1);
  }
  vector<int> myvec;
  cout << mx << endl;
  while (idx != -1) {
    myvec.push_back(idx), idx = par[idx];
  }
  for (int i = myvec.size() - 1; i >= 0; i--)
    printf("%d%c", myvec[i], " \n"[i == 0]);
}
