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;
using namespace std;
long long h[100010];
long long s[100010];
int n;
int niz[1000001];
int ST[(1000001 << 2)];
int LAST[(1000001 << 2)];
int last[100010];
void Update(int idx, int x, int val, int last, int left, int right) {
  if (left == right) {
    ST[idx] = val;
    LAST[idx] = last;
    return;
  }
  if (x <= (left + right) / 2)
    Update(2 * idx, x, val, last, left, (left + right) / 2);
  else
    Update(2 * idx + 1, x, val, last, (left + right) / 2 + 1, right);
  ST[idx] = max(ST[2 * idx], ST[2 * idx + 1]);
  if (ST[2 * idx] > ST[2 * idx + 1])
    LAST[idx] = LAST[2 * idx];
  else
    LAST[idx] = LAST[2 * idx + 1];
}
pair<int, int> Query(int idx, int l, int r, int left, int right) {
  if (l <= left && right <= r) return make_pair(ST[idx], LAST[idx]);
  pair<int, int> ret = make_pair(0, -1);
  if (l <= (left + right) / 2)
    ret = max(ret, Query(2 * idx, l, r, left, (left + right) / 2));
  if (r > (left + right) / 2)
    ret = max(ret, Query(2 * idx + 1, l, r, (left + right) / 2 + 1, right));
  return ret;
}
int main() {
  int d;
  scanf("%d%d", &n, &d);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &h[i]);
    s[i] = h[i];
  }
  sort(s, s + n);
  int start = lower_bound(s, s + n, h[0]) - s;
  memset(last, -1, sizeof last);
  memset(LAST, -1, sizeof LAST);
  Update(1, start, 1, 1, 0, n - 1);
  last[1] = -1;
  for (int i = 1; i < n; i++) {
    long long hcur = h[i];
    int low = upper_bound(s, s + n, hcur - d) - s;
    int hig = lower_bound(s, s + n, hcur + d) - s;
    low--;
    pair<int, int> x = make_pair(0, -1), y = make_pair(0, -1);
    if (low >= 0) x = Query(1, 0, low, 0, n - 1);
    if (hig < n) y = Query(1, hig, n - 1, 0, n - 1);
    int lst = -1;
    if (x > y)
      lst = x.second;
    else
      lst = y.second;
    int pos = lower_bound(s, s + n, h[i]) - s;
    last[i + 1] = lst;
    Update(1, pos, max(x.first, y.first) + 1, i + 1, 0, n - 1);
  }
  printf("%d\n", Query(1, 0, n - 1, 0, n - 1).first);
  int lst = Query(1, 0, n - 1, 0, n - 1).second;
  stack<int> x;
  while (lst != -1) {
    x.push(lst);
    lst = last[lst];
  }
  while (!x.empty()) {
    printf("%d ", x.top());
    x.pop();
  }
  printf("\n");
  return 0;
}
