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 n, i, j, d, max1, a[100500], tmp, c[100500];
pair<long long, long long> b[100500];
vector<long long> f;
long long Abs(long long x) {
  if (x < 0) return -x;
  return x;
}
int main() {
  cin >> n >> d;
  for (i = 0; i < n; i++) cin >> a[i];
  max1 = 1;
  b[1].first = b[1].second = a[0];
  c[0] = 1;
  for (i = 1; i < n; i++) {
    if (a[i] >= b[max1].first + d || a[i] <= b[max1].first - d ||
        a[i] >= b[max1].second + d || a[i] <= b[max1].second - d) {
      max1++;
      tmp = i + 1;
      b[max1].first = b[max1].second = a[i];
      c[i] = max1;
    } else if (max1 == 1 || a[i] >= b[max1 - 1].first + d ||
               a[i] <= b[max1 - 1].first - d ||
               a[i] >= b[max1 - 1].second + d ||
               a[i] <= b[max1 - 1].second - d) {
      if (a[i] > b[max1].second) b[max1].second = a[i], c[i] = max1;
      if (a[i] < b[max1].first) b[max1].first = a[i], c[i] = max1;
    } else {
      if (max1 < 2) continue;
      if (max1 <= 2 || a[i] >= b[max1 - 2].first + d ||
          a[i] <= b[max1 - 2].first - d || a[i] >= b[max1 - 2].second + d ||
          a[i] <= b[max1 - 2].second - d) {
        if (a[i] > b[max1 - 1].second)
          b[max1 - 1].second = a[i], c[i] = max1 - 1;
        if (a[i] < b[max1 - 1].first) b[max1 - 1].first = a[i], c[i] = max1 - 1;
      }
    }
  }
  cout << max1 << endl;
  f.push_back(tmp);
  long long start = tmp - 1;
  n = start;
  for (i = n; i >= 0; i--)
    if (c[i] == max1 - 1 && Abs(a[i] - a[start]) >= d) {
      f.push_back(i + 1);
      max1--;
      start = i;
    }
  reverse(f.begin(), f.end());
  for (i = 0; i < f.size(); i++) cout << f[i] << " ";
  return 0;
}
