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;
int tmin[100100], tmax[100100], res[100100];
int t;
long long b[100100], a[100100];
int n;
long long d;
int findmax(long long x) {
  int fir = 0, las = t + 1;
  while (fir + 1 < las) {
    int mid = (fir + las) >> 1;
    if (b[mid] >= x)
      las = mid;
    else
      fir = mid;
  }
  return las;
}
int findmin(long long x) {
  int fir = 0, las = t + 1;
  while (fir + 1 < las) {
    int mid = (fir + las) >> 1;
    if (b[mid] <= x)
      fir = mid;
    else
      las = mid;
  }
  return fir;
}
int main() {
  scanf("%d%I64d", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  t = 1;
  for (int i = 2; i <= n; i++)
    if (b[i] > b[t]) {
      t++;
      b[t] = b[i];
    }
  int mmax = 0;
  for (int i = n; i >= 1; i--) {
    res[i] = 1;
    long long h = a[i] + d;
    int x = findmax(h);
    while (x <= t) {
      res[i] = max(res[i], tmax[x] + 1);
      x += (x & (-x));
    }
    h = a[i] - d;
    x = findmin(h);
    while (x > 0) {
      res[i] = max(res[i], tmin[x] + 1);
      x -= (x & (-x));
    }
    x = findmax(a[i]);
    int y = x;
    while (x > 0) {
      tmax[x] = max(tmax[x], res[i]);
      x -= (x & (-x));
    }
    while (y <= t) {
      tmin[y] = max(tmin[y], res[i]);
      y += (y & (-y));
    }
    mmax = max(mmax, res[i]);
  }
  printf("%d\n", mmax);
  long long be = 100000000000000000;
  for (int i = 1; i <= n; i++)
    if (res[i] == mmax && abs(a[i] - be) >= d) {
      printf("%d ", i);
      mmax--;
      be = a[i];
      if (mmax == 0) break;
    }
}
