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;
int const N = 101010;
long long h[N];
struct node {
  long long h;
  int idx, c, lc, rc;
  friend bool operator<(node a, node b) { return a.h < b.h; }
} p[N];
bool cmp(node a, node b) { return a.idx < b.idx; }
int inv[N];
int pre[N];
struct tod {
  int val;
  int idx;
  friend bool operator<(tod a, tod b) { return a.val < b.val; }
} tmp, tree[N * 4];
void update(int l, int r, int a, int b, int p) {
  if (l == r) {
    tree[p].val = b, tree[p].idx = l;
    return;
  }
  int mid = (l + r) / 2, lp = p << 1, rp = lp | 1;
  if (a <= mid)
    update(l, mid, a, b, lp);
  else
    update(mid + 1, r, a, b, rp);
  tree[p] = max(tree[lp], tree[rp]);
}
tod query(int l, int r, int a, int b, int p) {
  if (l == a && r == b) {
    return tree[p];
  }
  int mid = (l + r) / 2, lp = p << 1, rp = lp | 1;
  if (b <= mid)
    return query(l, mid, a, b, lp);
  else if (a > mid)
    return query(mid + 1, r, a, b, rp);
  else
    return max(query(l, mid, a, mid, lp), query(mid + 1, r, mid + 1, b, rp));
}
vector<int> ans;
int main() {
  int n;
  long long d;
  scanf("%d%I64d", &n, &d);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &h[i]);
  if (d == 0) {
    printf("%d\n", n);
    for (int i = 1; i <= n; ++i) printf("%d ", i);
    puts("");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    p[i].h = h[i];
    p[i].idx = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i) {
    inv[i] = p[i].idx;
    p[i].c = i;
    if (i == 1)
      p[i].lc = 0;
    else {
      p[i].lc = p[i - 1].lc;
      while (p[i].h - d >= p[p[i].lc + 1].h) ++p[i].lc;
    }
  }
  for (int i = n; i >= 1; --i) {
    if (i == n)
      p[i].rc = i + 1;
    else {
      p[i].rc = p[i + 1].rc;
      while (p[i].h + d <= p[p[i].rc - 1].h) --p[i].rc;
    }
  }
  sort(p + 1, p + n + 1, cmp);
  int ret = 0, rk = 0;
  for (int i = 1; i <= n; ++i) {
    int lc = p[i].lc, rc = p[i].rc;
    tod t;
    t.idx = 0, t.val = 0;
    if (lc > 0) t = max(t, query(1, n, 1, lc, 1));
    if (rc < n) t = max(t, query(1, n, rc, n, 1));
    pre[i] = inv[t.idx];
    update(1, n, p[i].c, t.val + 1, 1);
    if (t.val + 1 > ret) {
      ret = t.val + 1;
      rk = i;
    }
  }
  for (int i = 0; i < ret; ++i) {
    ans.push_back(rk);
    rk = pre[rk];
  }
  printf("%d\n", ret);
  for (int i = ans.size() - 1; i >= 0; --i) {
    if (i == 0)
      printf("%d\n", ans[i]);
    else
      printf("%d ", ans[i]);
  }
  return 0;
}
