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;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
vector<long long int> a(100005);
vector<long long int> v(100005);
long long int d;
vector<pair<long long int, long long int>> tree(500005, {0, 0});
long long int n, m;
pair<long long int, long long int> combine(
    pair<long long int, long long int> a,
    pair<long long int, long long int> b) {
  if (a.first >= b.first) {
    return a;
  } else
    return b;
}
pair<long long int, long long int> sum(long long int tl, long long int tr,
                                       long long int l, long long int r,
                                       long long int v) {
  if (l > r) return {-1, -1};
  long long int mid = tl + (tr - tl) / 2;
  if (tl == l && tr == r) return tree[v];
  return combine(sum(tl, mid, l, min(mid, r), 2 * v),
                 sum(mid + 1, tr, max(mid + 1, l), r, 2 * v + 1));
}
void update(long long int l, long long int r, long long int pos,
            long long int val, long long int ind, long long int v) {
  if (l == r) {
    if (tree[v].first < val) {
      tree[v].first = val;
      tree[v].second = ind;
      a[pos] = val;
    }
  } else {
    long long int mid = l + (r - l) / 2;
    if (pos <= mid)
      update(l, mid, pos, val, ind, 2 * v);
    else
      update(mid + 1, r, pos, val, ind, 2 * v + 1);
    tree[v] = combine(tree[2 * v], tree[2 * v + 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> d;
  set<long long int> s;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    s.insert(v[i]);
  }
  long long int inx = -1;
  map<long long int, long long int> mp;
  long long int c = 1;
  for (auto i : s) mp[i] = c++;
  vector<long long int> dp(n, 0);
  vector<long long int> back(n, 0);
  long long int ma = 0;
  for (long long int i = 0; i < n; i++) {
    dp[i] = 1;
    back[i] = i;
    long long int a, b;
    auto ind = s.lower_bound(v[i] + d);
    if (ind != s.end()) {
      a = mp[*ind];
      pair<long long int, long long int> val = sum(1, c, a, c, 1);
      if (val.first + 1 > dp[i]) {
        dp[i] = 1 + val.first;
        back[i] = val.second;
      }
    }
    ind = s.upper_bound(v[i] - d);
    if (ind != s.begin()) {
      ind--;
      if (ind == s.end())
        b = c;
      else
        b = mp[*ind];
      pair<long long int, long long int> val = sum(1, c, 1, b, 1);
      if (val.first + 1 > dp[i]) {
        dp[i] = val.first + 1;
        back[i] = val.second;
      }
    }
    update(1, c, mp[v[i]], dp[i], i, 1);
    if (dp[i] > ma) inx = i;
    ma = max(ma, dp[i]);
  }
  vector<long long int> ans;
  ans.push_back(inx + 1);
  while (inx != back[inx]) {
    inx = back[inx];
    ans.push_back(inx + 1);
  }
  cout << ma << "\n";
  reverse((ans).begin(), (ans).end());
  for (auto i : ans) cout << i << " ";
}
