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;
long long height[200000], tree[600000], dp[200001], track[200000];
map<long long, long long> cheight;
set<long long> comp, temp;
void init(long long node, long long b, long long e) {
  if (b == e) {
    tree[node] = 100000;
    return;
  }
  long long Left = node * 2;
  long long Right = node * 2 + 1;
  long long mid = (b + e) / 2;
  init(Left, b, mid);
  init(Right, mid + 1, e);
  tree[node] = 100000;
}
void update(long long node, long long l, long long r, long long pos,
            long long u) {
  if (l == r) {
    tree[node] = u;
    return;
  }
  long long mid = (l + r) / 2;
  if (pos <= mid)
    update(2 * node, l, mid, pos, u);
  else
    update(2 * node + 1, mid + 1, r, pos, u);
  if (dp[tree[2 * node]] >= dp[tree[2 * node + 1]])
    tree[node] = tree[2 * node];
  else
    tree[node] = tree[2 * node + 1];
}
long long query(long long node, long long l, long long r, long long ql,
                long long qr) {
  if (ql > r || qr < l) return 100000;
  if (ql <= l && qr >= r) return tree[node];
  long long mid = (l + r) / 2;
  long long temp1 = query(2 * node, l, mid, ql, qr);
  long long temp2 = query(2 * node + 1, mid + 1, r, ql, qr);
  if (dp[temp1] > dp[temp2])
    return temp1;
  else
    return temp2;
}
int main() {
  long long i, j, k;
  long long n, d;
  long long pos, jump;
  scanf("%lld %lld", &n, &d);
  for (i = 0; i < n; i++) {
    scanf("%lld", &height[i]);
    comp.insert(height[i]);
  }
  i = 0;
  for (auto itr = comp.begin(); itr != comp.end(); itr++, i++)
    cheight[*itr] = i;
  init(1, 0, n - 1);
  dp[n - 1] = 1;
  update(1, 0, n - 1, cheight[height[n - 1]], n - 1);
  temp.insert(height[n - 1]);
  for (i = 0; i < n; i++) track[i] = -1;
  for (i = n - 2; i >= 0; i--) {
    dp[i] = 1;
    auto itr = temp.upper_bound(height[i] - d);
    if (itr != temp.begin()) {
      itr--;
      j = cheight[*itr];
      k = query(1, 0, n - 1, 0, j);
      dp[i] = 1 + dp[k];
      track[i] = k;
    }
    itr = temp.lower_bound(height[i] + d);
    if (itr != temp.end()) {
      j = cheight[*itr];
      k = query(1, 0, n - 1, j, n - 1);
      if ((1 + dp[k]) > dp[i]) {
        dp[i] = 1 + dp[k];
        track[i] = k;
      }
    }
    temp.insert(height[i]);
    update(1, 0, n - 1, cheight[height[i]], i);
  }
  for (i = 0, jump = 0, pos = 0; i < n; i++) {
    if (dp[i] > jump) {
      jump = dp[i];
      pos = i;
    }
  }
  printf("%lld\n", jump);
  i = pos;
  while (i >= 0) {
    printf("%lld ", i + 1);
    i = track[i];
  }
}
