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 n, d;
long long v[100001], r[100001], pre[100001];
void lis() {
  int iMax = 0, iPos;
  for (int i = 0; i < n; i++) {
    for (int j = (i > 500 ? i - 500 : 0); j < i; j++) {
      if (abs(v[i] - v[j]) >= d && r[j] + 1 > r[i]) {
        r[i] = r[j] + 1;
        pre[i] = j;
      }
      if (r[i] > iMax) {
        iPos = i;
        iMax = r[i];
      }
    }
  }
  stack<int> nums;
  while (iPos >= 0) {
    nums.push(iPos + 1);
    iPos = pre[iPos];
  }
  cout << iMax + 1 << "\n";
  while (!nums.empty()) {
    cout << nums.top() << " ";
    nums.pop();
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> d;
  for (int i = 0; i < n; i++) cin >> v[i];
  fill(pre, pre + n + 1, -1);
  lis();
  return 0;
}
