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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, 1, -1};
int n;
long long d;
int sz;
vector<long long> xs;
long long a[100010];
int p[100010];
int pre[100010];
map<int, int> m;
struct segtree_max {
  vector<pair<long long, int> > dat;
  void init(int n_) {
    sz = 1;
    while (sz < n_) sz *= 2;
    dat.resize(sz * 2);
    for (int i = sz - 1; i <= sz * 2 - 2; ++i)
      dat[i] = make_pair(LLONG_MIN, i + 1 - sz);
    for (int i = sz - 2; i >= 0; --i)
      dat[i] = max(dat[i * 2 + 1], dat[i * 2 + 2]);
  }
  void update(int pos, long long x) {
    pos += sz - 1;
    dat[pos].first = x;
    while (pos > 0) {
      pos = (pos - 1) / 2;
      dat[pos] = max(dat[pos * 2 + 1], dat[pos * 2 + 2]);
    }
  }
  pair<long long, int> get(int a, int b, int k = 0, int l = 0, int r = sz) {
    if (b <= l || r <= a) return make_pair(LLONG_MIN, -1);
    if (a <= l && r <= b) return dat[k];
    return max(get(a, b, k * 2 + 1, l, (l + r) / 2),
               get(a, b, k * 2 + 2, (l + r) / 2, r));
  }
} seg;
int main() {
  scanf("%d %I64d", &n, &d);
  xs.resize(n * 3);
  for (int i = (0); i < (n); ++i) {
    scanf("%I64d", &a[i]);
    xs[i * 3] = a[i];
    xs[i * 3 + 1] = a[i] - d;
    xs[i * 3 + 2] = a[i] + d;
  }
  sort(xs.begin(), xs.end());
  xs.erase(unique(xs.begin(), xs.end()), xs.end());
  for (int i = (0); i < (n); ++i)
    p[i] = lower_bound(xs.begin(), xs.end(), a[i]) - xs.begin();
  seg.init(xs.size());
  seg.update(p[0], 1);
  pre[0] = -1;
  m[p[0]] = 0;
  for (int i = 1; i < n; ++i) {
    long long lo = a[i] - d, hi = a[i] + d;
    lo = upper_bound(xs.begin(), xs.end(), lo) - xs.begin();
    hi = lower_bound(xs.begin(), xs.end(), hi) - xs.begin();
    pair<long long, int> s = make_pair(LLONG_MIN, -1),
                         t = make_pair(LLONG_MIN, -1);
    s = seg.get(0, lo);
    t = seg.get(hi, xs.size());
    s = max(s, t);
    if (s.first != LLONG_MIN) {
      pre[i] = m[s.second];
      seg.update(p[i], s.first + 1);
    } else {
      pre[i] = -1;
      seg.update(p[i], 1);
    }
    m[p[i]] = i;
  }
  pair<long long, int> s = seg.get(0, xs.size());
  printf("%I64d\n", s.first);
  int cur = m[s.second];
  vector<int> ans(s.first);
  for (int i = (0); i < (s.first); ++i) {
    ans.push_back(cur + 1);
    cur = pre[cur];
  }
  reverse(ans.begin(), ans.end());
  for (int i = (0); i < (s.first); ++i)
    printf("%d%c", ans[i], i != s.first - 1 ? ' ' : '\n');
  return 0;
}
