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>
const int maxn = 100010;
int a[maxn];
std::pair<long long, int> p[maxn];
struct seg {
  int l, r;
  int o, m;
} t[maxn << 2];
void inline pushup(int id) {
  t[id].m = std::max(t[(id << 1)].m, t[(id << 1 | 1)].m);
  t[id].o =
      t[(id << 1)].m > t[(id << 1 | 1)].m ? t[(id << 1)].o : t[(id << 1 | 1)].o;
}
void build(int id, int l, int r) {
  t[id].l = l, t[id].r = r;
  if (l == r) return (void)(t[id].o = l, t[id].m = 0);
  int mid = (l + r) >> 1;
  build((id << 1), l, mid), build((id << 1 | 1), mid + 1, r);
  pushup(id);
}
void ins(int id, int p, int v) {
  if (t[id].l == t[id].r) return (void)(t[id].m = v);
  int mid = (t[id].l + t[id].r) >> 1;
  ins(p <= mid ? (id << 1) : (id << 1 | 1), p, v);
  pushup(id);
}
std::pair<int, int> max(int id, int l, int r) {
  if (l > r) return std::make_pair(0, 0);
  if (t[id].l == l && t[id].r == r) return std::make_pair(t[id].m, t[id].o);
  int mid = (t[id].l + t[id].r) >> 1;
  if (r <= mid)
    return max((id << 1), l, r);
  else if (l > mid)
    return max((id << 1 | 1), l, r);
  else
    return std::max(max((id << 1), l, mid), max((id << 1 | 1), mid + 1, r));
}
long long h[maxn];
int l[maxn], r[maxn], b[maxn], g[maxn], f[maxn], q[maxn];
void dfs(int u) {
  if (b[u]) dfs(b[u]);
  printf("%d ", u);
}
int main() {
  static int n, d, ans, pos, lp = 1, rp = 1;
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; ++i) scanf("%I64d", h + i);
  for (int i = 1; i <= n; ++i) p[i] = std::make_pair(h[i], i);
  std::sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i) g[p[i].second] = i;
  for (int i = 1; i <= n; ++i) q[g[i]] = i;
  for (int i = 1; i <= n; ++i) {
    while (rp < n && p[rp + 1].first - p[i].first < d) ++rp;
    while (lp < n && p[i].first - p[lp].first >= d) ++lp;
    l[p[i].second] = lp, r[p[i].second] = rp;
  }
  build(1, 1, n);
  for (int i = 1; i <= n; ++i) {
    std::pair<int, int> u = std::max(max(1, 1, l[i] - 1), max(1, r[i] + 1, n));
    f[i] = u.first + 1;
    if (u.first) b[i] = q[u.second];
    ins(1, g[i], f[i]);
  }
  for (int i = 1; i <= n; ++i)
    if (f[i] > ans) ans = f[i], pos = i;
  printf("%d\n", ans);
  dfs(pos);
  return 0;
}
