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 oo = 1e9 + 7;
struct SegmentTree {
  long long n;
  vector<pair<long long, long long> > st;
  SegmentTree() {}
  SegmentTree(long long _n) {
    this->n = _n;
    st.assign(4 * n + 5, {-oo, 0});
  }
  long long Left(long long x) { return (x << 1); }
  long long Right(long long x) { return (x << 1) + 1; }
  void update(long long id, long long L, long long R, long long i,
              long long val) {
    if (L > i || R < i) return;
    if (L == R) {
      st[id] = {val, i};
      return;
    }
    long long mid = (L + R) >> 1;
    update(Left(id), L, mid, i, val);
    update(Right(id), mid + 1, R, i, val);
    st[id] = max(st[Left(id)], st[Right(id)]);
  }
  pair<long long, long long> get(long long id, long long L, long long R,
                                 long long i, long long j) {
    if (L > j || R < i) return {0, 0};
    if (L >= i && R <= j) return st[id];
    long long mid = (L + R) >> 1;
    return max(get(Left(id), L, mid, i, j), get(Right(id), mid + 1, R, i, j));
  }
  void update(long long i, long long val) { update(1, 1, n, i, val); }
  pair<long long, long long> get(long long l, long long r) {
    if (l > r) return {0, 0};
    return get(1, 1, n, l, r);
  }
};
const long long N = 1e5 + 5;
long long n, a[N], l[N], r[N], d, trace[N], pos[N], dp[N];
vector<pair<long long, long long> > A;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> d;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    A.push_back({a[i], i});
  }
  A.push_back({-1, -1});
  sort(A.begin(), A.end());
  SegmentTree IT = SegmentTree(n);
  long long j = 1;
  for (long long i = 1; i <= n; ++i) {
    pos[A[i].second] = i;
    while (A[i].first - A[j].first >= d && j < i) {
      r[A[j].second] = i - 1;
      j++;
    }
    l[A[i].second] = j;
  }
  for (long long i = 1; i <= n; ++i) {
    if (r[i] == 0) r[i] = n + 1;
    pair<long long, long long> V =
        max(IT.get(1, l[i] - 1), IT.get(r[i] + 1, n));
    V = max(V, {0, 0});
    if (V.first == 0) {
      trace[i] = 0;
      dp[i] = 1;
      IT.update(pos[i], 1);
      continue;
    }
    trace[i] = A[V.second].second;
    dp[i] = V.first + 1;
    IT.update(pos[i], V.first + 1);
  }
  long long res = 0, id = -1;
  for (long long i = 1; i <= n; ++i)
    if (dp[i] > res) {
      res = dp[i];
      id = i;
    }
  cout << res << endl;
  vector<long long> O;
  O.clear();
  while (id) {
    O.push_back(id);
    id = trace[id];
  }
  reverse(O.begin(), O.end());
  for (long long v : O) cout << v << ' ';
  return 0;
}
