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 N = 3e5 + 5;
set<long long> s;
set<long long>::iterator it;
long long a[N], d, n, h[N], ans = 0, last = 0;
pair<long long, long long> dp[N];
struct IT {
  pair<long long, long long> s[4 * N];
  void init() { memset(s, 0, sizeof(s)); }
  void update(long long node, long long l, long long r, long long p,
              long long q, long long pos, pair<long long, long long> val) {
    if (l > q || r < p) return;
    if (l == r) {
      s[node] = max(s[node], val);
      return;
    }
    long long mid = (l + r) >> 1;
    if (mid >= pos)
      update(2 * node, l, mid, p, q, pos, val);
    else
      update(2 * node + 1, mid + 1, r, p, q, pos, val);
    s[node] = max(s[2 * node], s[2 * node + 1]);
  }
  pair<long long, long long> get(long long node, long long l, long long r,
                                 long long p, long long q) {
    if (l > q || r < p) return pair<long long, long long>(0, 0);
    if (l >= p && r <= q) return s[node];
    long long mid = (l + r) >> 1;
    pair<long long, long long> curr =
        max(get(2 * node, l, mid, p, q), get(2 * node + 1, mid + 1, r, p, q));
    return curr;
  }
} T;
int main() {
  scanf("%lld%lld", &n, &d);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    s.insert(a[i]);
    s.insert(a[i] - d);
    s.insert(a[i] + d);
  }
  long long cnt = 0;
  for (it = s.begin(); it != s.end(); it++) h[++cnt] = *it;
  T.init();
  memset(dp, 0, sizeof(dp));
  for (long long i = 1; i <= n; ++i) {
    long long cs1 = lower_bound(h + 1, h + cnt + 1, a[i]) - h;
    long long cs2 = lower_bound(h + 1, h + cnt + 1, a[i] - d) - h;
    long long cs3 = lower_bound(h + 1, h + cnt + 1, a[i] + d) - h;
    pair<long long, long long> curr =
        max(T.get(1, 1, cnt, 1, cs2), T.get(1, 1, cnt, cs3, cnt));
    dp[i] = pair<long long, long long>(curr.first + 1, curr.second);
    T.update(1, 1, cnt, 1, cnt, cs1,
             pair<long long, long long>(dp[i].first, i));
    if (dp[i].first > ans) {
      ans = dp[i].first;
      last = i;
    }
  }
  cout << ans << "\n";
  vector<long long> res;
  res.clear();
  while (dp[last].second) {
    res.push_back(last);
    last = dp[last].second;
  }
  res.push_back(last);
  reverse(res.begin(), res.end());
  for (long long i = 0; i < res.size(); ++i) cout << res[i] << " ";
}
