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 long long top = (long long)1e15;
struct Seg {
  Seg *l, *r;
  int p;
  Seg() { l = r = NULL, p = 0; }
} * root;
int n, ans, mx, from[100005], t, f[100005], point;
long long x, d;
inline int Max(int x, int y) {
  if (f[x] > f[y]) return x;
  return y;
}
int query(Seg *x, long long l, long long r, long long a, long long b) {
  if (b < a) return 0;
  if (!x) return 0;
  if (l == a && b == r) return x->p;
  long long mid = l + r >> 1;
  if (b <= mid) return query(x->l, l, mid, a, b);
  if (a > mid) return query(x->r, mid + 1, r, a, b);
  return Max(query(x->l, l, mid, a, mid), query(x->r, mid + 1, r, mid + 1, b));
}
void insert(Seg *&x, long long l, long long r, long long k) {
  if (!x) x = new Seg();
  if (l == r) {
    x->p = point;
    return;
  }
  long long mid = l + r >> 1;
  if (mid >= k)
    insert(x->l, l, mid, k);
  else
    insert(x->r, mid + 1, r, k);
  x->p = Max(x->l ? x->l->p : 0, x->r ? x->r->p : 0);
}
void dfs(int p) {
  if (!p) return;
  dfs(from[p]);
  printf("%d ", p);
}
int main() {
  scanf("%d%I64d", &n, &d);
  root = new Seg();
  for (int i = 1; i <= n; ++i) {
    point = i;
    scanf("%I64d", &x);
    mx = query(root, 1, top, 1, x - d);
    mx = Max(mx, query(root, 1, top, x + d, top));
    f[i] = f[mx] + 1;
    insert(root, 1, top, x);
    if (f[i] > ans) {
      ans = f[i];
      t = i;
    }
    from[i] = mx;
  }
  printf("%d\n", ans);
  dfs(t);
}
