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 = 100500;
struct node {
  node *l, *r;
  long long x;
  int id, mx, y;
  int _id, _mx;
  node() {
    l = r = NULL;
    x = 0;
    id = mx = y = 0;
    _mx = _id = 0;
  }
} * t;
void upd(node *&x) {
  if (!x) return;
  x->mx = x->_mx;
  x->id = x->_id;
  if (x->l) {
    if (x->mx < x->l->mx) x->mx = x->l->mx, x->id = x->l->id;
  }
  if (x->r) {
    if (x->mx < x->r->mx) x->mx = x->r->mx, x->id = x->r->id;
  }
}
node *merge(node *l, node *r) {
  if (!l) return r;
  if (!r) return l;
  if (l->y > r->y) {
    l->r = merge(l->r, r);
    upd(l);
    return l;
  } else {
    r->l = merge(l, r->l);
    upd(r);
    return r;
  }
}
void split(node *t, long long x, node *&l, node *&r) {
  if (!t) {
    l = r = NULL;
    return;
  }
  if (t->x < x) {
    l = t;
    split(l->r, x, l->r, r);
    upd(l);
  } else {
    r = t;
    split(r->l, x, l, r->l);
    upd(r);
  }
}
void add(long long x, int mx, int id) {
  node *z = new node;
  z->x = x;
  z->y = rand();
  z->_mx = z->mx = mx;
  z->_id = z->id = id;
  node *l = NULL, *r = NULL;
  split(t, x, l, r);
  t = merge(l, merge(z, r));
}
void out(node *&x) {
  if (!x) return;
  out(x->l);
  cout << x->x << " ";
  out(x->r);
}
int n, d, dp[N], p[N];
long long a[N];
int main() {
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 1; i <= n; ++i) {
    dp[i] = 1;
    node *L = NULL, *R = NULL, *M = NULL;
    split(t, a[i] - d + 1, L, R);
    split(R, a[i] + d, M, R);
    if (L) {
      if (L->mx + 1 > dp[i]) dp[i] = L->mx + 1, p[i] = L->id;
    }
    if (R) {
      if (R->mx + 1 > dp[i]) dp[i] = R->mx + 1, p[i] = R->id;
    }
    t = merge(L, merge(M, R));
    add(a[i], dp[i], i);
  }
  int mx = 0, id = 0;
  for (int i = 1; i <= n; ++i) {
    if (dp[i] > mx) mx = dp[i], id = i;
  }
  printf("%d\n", mx);
  vector<int> ans;
  while (id) {
    ans.push_back(id);
    id = p[id];
  }
  for (int i = ans.size() - 1; i >= 0; --i) printf("%d ", ans[i]);
  return 0;
}
