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 len[200000], pos[200000];
long long heights[200000];
int n, d;
vector<int> v;
int main() {
  cin >> n >> d;
  for (int i = 1; i <= n; i++) cin >> heights[i], len[i] = 1, pos[i] = i;
  for (int i = 1; i <= n; i++) {
    for (int j = max(1, i - 700); j < i; j++)
      if (abs(heights[i] - heights[j]) >= d && len[j] + 1 >= len[i])
        len[i] = len[j] + 1, pos[i] = j;
  }
  int Max = -1, in;
  for (int i = 1; i <= n; i++)
    if (Max < len[i]) Max = len[i], in = i;
  cout << len[in] << endl;
  while (in != pos[in]) {
    v.push_back(in);
    in = pos[in];
  }
  v.push_back(pos[in]);
  for (int i = v.size() - 1; i >= 0; i--) cout << v[i] << " ";
}
