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;
pair<int, int> tree[1 << 18];
long long H[1 << 18], h[1 << 18];
int dp[1 << 18], pre[1 << 18];
void build(int i, int st, int en) {
  tree[i] = make_pair(-1, 0);
  if (st == en) return;
  int mid = (st + en) >> 1, lt = i << 1, rt = lt + 1;
  build(lt, st, mid);
  build(rt, mid + 1, en);
  return;
}
void modify(int i, int st, int en, int id, int len) {
  if (st == en) {
    if (tree[i].second < len) tree[i] = make_pair(id, len);
    return;
  }
  int mid = (st + en) >> 1, lt = i << 1, rt = lt + 1;
  if (h[id] <= H[mid])
    modify(lt, st, mid, id, len);
  else
    modify(rt, mid + 1, en, id, len);
  if (tree[lt].second > tree[rt].second)
    tree[i] = tree[lt];
  else
    tree[i] = tree[rt];
  return;
}
pair<int, int> query(int i, int st, int en, long long x, long long y) {
  if (y < x) return make_pair(-1, 0);
  if (x > H[en] || y < H[st]) return make_pair(-1, 0);
  if (x <= H[st] && H[en] <= y) return tree[i];
  int mid = (st + en) >> 1, lt = i << 1, rt = lt + 1;
  pair<int, int> a = query(lt, st, mid, x, y);
  pair<int, int> b = query(rt, mid + 1, en, x, y);
  if (a.second > b.second) return a;
  return b;
}
int main() {
  int i, j, k, test_case, caseno = 1, n;
  long long d;
  scanf("%d %I64d", &n, &d);
  for (i = 0; i < n; i++) {
    scanf("%I64d", &h[i]);
    H[i] = h[i];
  }
  sort(H, H + n);
  int N = (unique(H, H + n) - H);
  build(1, 0, N - 1);
  for (i = 0; i < n; i++) {
    pair<int, int> a = query(1, 0, N - 1, H[0], h[i] - d);
    pair<int, int> b = query(1, 0, N - 1, h[i] + d, H[N - 1]);
    if (a.second < b.second) a = b;
    dp[i] = a.second + 1;
    pre[i] = a.first;
    modify(1, 0, N - 1, i, dp[i]);
  }
  int ans = 0;
  for (i = 0; i < n; i++) {
    if (dp[i] > dp[ans]) ans = i;
  }
  printf("%d\n", dp[ans]);
  stack<int> s;
  while (ans != -1) {
    s.push(ans + 1);
    ans = pre[ans];
  }
  printf("%d", s.top());
  s.pop();
  while (s.empty() == 0) {
    printf(" %d", s.top());
    s.pop();
  }
  printf("\n");
}
