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 in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[1000010];
  scanf("%s", ch);
  return ch;
}
const int MAX_LG = 21;
const long long maxn = 2e6 + 10;
const long long base = 29;
const long long mod = 1e9 + 7;
const long long INF = 1e9 + 100;
pair<long long, long long> a[maxn];
long long cmp[maxn], fen[maxn][2];
long long dp[maxn], pre[maxn];
long long id;
long long res;
inline pair<long long, long long> get(long long l, long long r) {
  long long ret = 0, id = -1;
  for (l += 2; l; l -= l & -l) {
    if (fen[l][0] != -1 && dp[fen[l][0]] > ret) {
      ret = dp[fen[l][0]];
      id = fen[l][0];
    }
  }
  for (r = maxn - r - 1; r; r -= r & -r) {
    if (fen[r][1] != -1 && dp[fen[r][1]] > ret) {
      ret = dp[fen[r][1]], id = fen[r][1];
    }
  }
  return {ret, id};
}
inline void update(long long p) {
  long long value = dp[p], saveid = p;
  for (long long p = saveid + 2; p < maxn; p += p & -p) {
    if (fen[p][0] == -1 || dp[fen[p][0]] < value) fen[p][0] = saveid;
  }
  for (long long p = maxn - saveid - 1; p < maxn; p += p & -p) {
    if (fen[p][1] == -1 || dp[fen[p][1]] < value) fen[p][1] = saveid;
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, d;
  cin >> n >> d;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first, a[i].second = i;
  }
  for (long long i = 0; i < maxn; i++) fen[i][0] = fen[i][1] = -1;
  sort(a, a + n);
  for (long long i = 0; i < n; i++) cmp[a[i].second] = i;
  for (long long i = 0; i < n; i++) {
    long long c = cmp[i];
    pair<long long, long long> rs =
        get(lower_bound(a, a + n, make_pair(a[c].first - d, n)) - a - 1,
            lower_bound(a, a + n, make_pair(a[c].first + d, -1ll)) - a);
    dp[c] = ++rs.first;
    pre[c] = rs.second;
    update(c);
  }
  res = *max_element(dp, dp + n);
  for (long long i = 0; i < n; i++)
    if (res == dp[i]) id = i;
  cout << res << "\n";
  vector<long long> vc;
  while (id != -1) vc.push_back(a[id].second), id = pre[id];
  sort(vc.begin(), vc.end());
  for (auto v : vc) cout << v + 1 << " ";
}
