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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int N = 100005;
int n, d;
long long h[N];
long long hh[N];
int z;
map<long long, int> mp;
pair<int, int> t[N * 4];
pair<int, int> mer(const pair<int, int>& l, const pair<int, int>& r) {
  if (l.first > r.first) return l;
  return r;
}
void ubd(int tl, int tr, int x, pair<int, int> y, int pos) {
  if (tl == tr) {
    t[pos] = mer(t[pos], y);
    return;
  }
  int m = (tl + tr) / 2;
  if (x <= m)
    ubd(tl, m, x, y, pos * 2);
  else
    ubd(m + 1, tr, x, y, pos * 2 + 1);
  t[pos] = mer(t[pos * 2], t[pos * 2 + 1]);
}
pair<int, int> qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return make_pair(0, 0);
  if (tl == l && tr == r) return t[pos];
  int m = (tl + tr) / 2;
  return mer(qry(tl, m, l, min(m, r), pos * 2),
             qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
int dp[N];
int p[N];
int main() {
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &h[i]);
    hh[i] = h[i];
  }
  sort(hh + 1, hh + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (hh[i] != hh[i - 1]) mp[hh[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    int l = 1, r = n;
    int ll = n + 1;
    while (l <= r) {
      int m = (l + r) / 2;
      if (hh[m] >= h[i] + d) {
        ll = m;
        r = m - 1;
      } else
        l = m + 1;
    }
    l = 1, r = n;
    int rr = 0;
    while (l <= r) {
      int m = (l + r) / 2;
      if (hh[m] <= h[i] - d) {
        rr = m;
        l = m + 1;
      } else
        r = m - 1;
    }
    pair<int, int> u = mer(qry(1, n, 1, rr, 1), qry(1, n, ll, n, 1));
    dp[i] = u.first + 1;
    p[i] = u.second;
    ubd(1, n, mp[h[i]], make_pair(dp[i], i), 1);
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
  for (int i = 1; i <= n; ++i) {
    if (dp[i] == ans) {
      vector<int> ans;
      while (i) {
        ans.push_back(i);
        i = p[i];
      }
      reverse(ans.begin(), ans.end());
      for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
      printf("\n");
      return 0;
    }
  }
  return 0;
}
