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;
const int mx = 1e5 + 5;
pair<int, int> tree[3 * mx];
int DP[mx], path[mx], rev[mx];
long long ara[mx], num[mx];
pair<long long, int> temp[mx];
void update(int pos, int L, int R, int node, int val) {
  if (L > node || R < node) return;
  if (L == R && L == node) {
    tree[pos].first = val;
    tree[pos].second = L;
    return;
  }
  int mid = (L + R) / 2;
  int x = pos * 2;
  update(x, L, mid, node, val);
  update(x + 1, mid + 1, R, node, val);
  if (tree[x].first > tree[x + 1].first)
    tree[pos] = tree[x];
  else
    tree[pos] = tree[x + 1];
}
pair<int, int> query(int pos, int L, int R, int l, int r) {
  if (L > r || l > R) return make_pair(0, 0);
  if (l <= L && R <= r) return tree[pos];
  int mid = (L + R) / 2, x = pos * 2;
  pair<int, int> y = query(x, L, mid, l, r);
  pair<int, int> z = query(x + 1, mid + 1, R, l, r);
  return max(y, z);
}
int main() {
  int i, n, d;
  pair<int, int> ans = make_pair(-1, -1);
  scanf("%d %d", &n, &d);
  for (i = 0; i < n; i++) {
    scanf("%lld", &ara[i]);
    temp[i].first = ara[i];
    temp[i].second = i;
  }
  sort(temp, temp + n);
  for (i = 0; i < n; i++) {
    rev[temp[i].second] = i;
    num[i] = temp[i].first;
  }
  for (i = n; i > 0; i--) {
    long long h1 = ara[i - 1] + d, h2 = ara[i - 1] - d;
    int x = lower_bound(num, num + n, h1) - num;
    int y = upper_bound(num, num + n, h2) - num;
    pair<int, int> A = query(1, 1, n, x + 1, n);
    pair<int, int> B = query(1, 1, n, 1, y);
    A = max(A, B);
    DP[i] = 1 + A.first;
    if (A.second > 0)
      path[i] = temp[A.second - 1].second + 1;
    else
      path[i] = -1;
    update(1, 1, n, rev[i - 1] + 1, DP[i]);
    ans = max(ans, make_pair(DP[i], i));
  }
  printf("%d\n", ans.first);
  int u = ans.second;
  while (u != -1) {
    printf("%d ", u);
    u = path[u];
  }
  return 0;
}
