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>
#pragma warning(disable : 4996)
using namespace std;
const int IT_MAX = 131072;
const int MOD = 1000000007;
const int INF = 2034567891;
const long long LL_INF = 1234567890123456789ll;
long long in[100050];
pair<long long, int> sor[100050];
int ch[100050];
int dp[100050];
int rev[100050];
pair<int, int> it[2 * IT_MAX + 50];
vector<int> ANS;
void update(int pos, pair<int, int> val) {
  it[pos] = val;
  while (pos > 1) {
    pos /= 2;
    it[pos] = max(it[pos * 2], it[pos * 2 + 1]);
  }
}
pair<int, int> getmx(int p1, int p2) {
  if (p1 > p2) return pair<int, int>(0, 0);
  pair<int, int> t = pair<int, int>(0, 0);
  if (p1 == p2) return it[p1];
  if (p1 % 2 == 1) {
    t = it[p1];
    p1++;
  }
  if (p1 == p2) return max(t, it[p1]);
  if (p2 % 2 == 0) {
    t = max(t, it[p2]);
    p2--;
  }
  return max(t, getmx(p1 / 2, p2 / 2));
}
int main() {
  int N, i;
  long long D;
  scanf("%d %I64d", &N, &D);
  for (i = 1; i <= N; i++) {
    scanf("%I64d", &in[i]);
    sor[i] = pair<long long, int>(in[i], i);
  }
  sort(sor + 1, sor + N + 1);
  for (i = 1; i <= N; i++) ch[sor[i].second] = i;
  for (i = 1; i <= N; i++) {
    int st = 1, en = N, mi, rv = 0;
    while (st <= en) {
      mi = (st + en) / 2;
      if (sor[mi].first <= in[i] - D) {
        rv = mi;
        st = mi + 1;
      } else
        en = mi - 1;
    }
    pair<int, int> ans = getmx(IT_MAX, IT_MAX + rv - 1);
    st = 1, en = N, mi, rv = N + 1;
    while (st <= en) {
      mi = (st + en) / 2;
      if (sor[mi].first >= in[i] + D) {
        rv = mi;
        en = mi - 1;
      } else
        st = mi + 1;
    }
    ans = max(ans, getmx(IT_MAX + rv - 1, IT_MAX + N - 1));
    dp[i] = ans.first + 1;
    rev[i] = ans.second;
    update(IT_MAX + ch[i] - 1, pair<int, int>(dp[i], i));
  }
  int p = 1;
  for (i = 2; i <= N; i++)
    if (dp[p] < dp[i]) p = i;
  printf("%d\n", dp[p]);
  while (p != 0) {
    ANS.push_back(p);
    p = rev[p];
  }
  for (i = ANS.size() - 1; i >= 0; i--) printf("%d ", ANS[i]);
  return 0;
}
