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 NMAX = 1e5 + 7;
int n, d;
long long h[NMAX], tree[4 * NMAX], dp[NMAX];
vector<long long> a, sol;
long long kkk(long long x) { return (x > 0) ? x : -x; }
void update(int node, int st, int en, int id, long long val) {
  if (st > en || id > en || id < st) return;
  if (st == en) {
    tree[node] = max(tree[node], val);
    return;
  }
  int mid = (st + en) / 2;
  update(2 * node, st, mid, id, val);
  update(2 * node + 1, mid + 1, en, id, val);
  tree[node] = max(tree[2 * node], tree[2 * node + 1]);
}
long long query(int node, int st, int en, int l, int r) {
  if (st > r || en < l || st > en) return -1;
  if (st >= l && en <= r) return tree[node];
  int mid = (st + en) / 2;
  return max(query(2 * node, st, mid, l, r),
             query(2 * node + 1, mid + 1, en, l, r));
}
int main() {
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &h[i]);
    a.push_back(h[i]);
  }
  sort(a.begin(), a.end());
  long long ans = -1;
  for (int i = n; i >= 1; i--) {
    long long x = lower_bound(a.begin(), a.end(), h[i] + d) - a.begin();
    long long tmp1 = query(1, 1, n, x + 1, n);
    dp[i] = tmp1 + 1;
    long long y = upper_bound(a.begin(), a.end(), h[i] - d) - a.begin();
    long long tmp2 = query(1, 1, n, 1, y);
    dp[i] = max(dp[i], tmp2 + 1);
    long long z = lower_bound(a.begin(), a.end(), h[i]) - a.begin();
    update(1, 1, n, z + 1, dp[i]);
    ans = max(ans, dp[i]);
  }
  int j = 0;
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    if (dp[i] == ans - sol.size() && (kkk(h[i] - h[j]) >= d || j == 0)) {
      sol.push_back(i);
      j = i;
    }
  }
  for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
  return 0;
}
