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 double eps = 1e-10;
const int N = 1e5 + 10;
const int M = 1e9 + 7;
long long a[N];
pair<long long, int> b[N];
int t[N], p[N];
struct IntervalTree {
  long long a[N << 2];
  int p;
  int qL, qR;
  void che(int o) {
    if (t[a[o << 1]] >= t[a[o << 1 | 1]])
      a[o] = a[o << 1];
    else
      a[o] = a[o << 1 | 1];
  }
  void build() { memset(a, 0, sizeof(a)); }
  void update(int L, int R, int o) {
    int M = L + (R - L) / 2;
    if (L == R) {
      a[o] = L;
    } else {
      if (p <= M)
        update(L, M, o << 1);
      else
        update(M + 1, R, o << 1 | 1);
      che(o);
    }
  }
  int query(int L, int R, int o) {
    int M = L + (R - L) / 2, ans = 0, se;
    if (qL <= L && R <= qR) return a[o];
    if (qL <= M) {
      se = query(L, M, o << 1);
      if (t[se] > t[ans]) ans = se;
    }
    if (M + 1 <= qR) {
      se = query(M + 1, R, o << 1 | 1);
      if (t[se] > t[ans]) ans = se;
    }
    return ans;
  }
} A;
int main() {
  long long n, d;
  scanf("%lld %lld", &n, &d);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    b[i].first = a[i];
    b[i].second = i + 1;
  }
  sort(b, b + n);
  long long L = b[0].first, R = b[n - 1].first;
  int Mx = 0, nod = 0;
  A.build();
  memset(t, 0, sizeof(t));
  for (int i = n - 1; i >= 0; i--) {
    long long li = a[i] - d, ri = a[i] + d;
    pair<long long, int> tt = make_pair(li, n + 1);
    int l = upper_bound(b, b + n, tt) - b;
    tt = make_pair(ri, -1);
    int r = lower_bound(b, b + n, tt) - b + 1;
    tt = make_pair(a[i], i + 1);
    int m = lower_bound(b, b + n, tt) - b + 1;
    int t1 = 0, t2 = 0, ti = 0;
    if (li >= L) {
      A.qL = 1;
      A.qR = l;
      t1 = A.query(1, n, 1);
    }
    if (ri <= R) {
      A.qL = r;
      A.qR = n;
      t2 = A.query(1, n, 1);
    }
    ti = t1;
    if (t[t2] > t[t1]) ti = t2;
    t[m] = t[ti] + 1;
    if (ti == 0)
      p[i] = 0;
    else
      p[i] = b[ti - 1].second;
    if (t[m] > Mx) {
      Mx = t[m];
      nod = i;
    }
    A.p = m;
    A.update(1, n, 1);
  }
  printf("%d\n", Mx);
  for (int i = nod + 1; i; i = p[i - 1]) printf("%d ", i);
  puts("");
  return 0;
}
