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 long long MAXN = 100 * 1000 + 23,
                MAXX = 1000ll * 1000 * 1000 * 1000 * 1000;
long long n, d, h[MAXN], ans[MAXN], par[MAXN];
struct node {
  pair<long long, long long> val{0, -1};
  node *lc, *rc;
} *rt = new node();
inline void qwe(node *v) {
  if (!(v->lc)) v->lc = new node();
  if (!(v->rc)) v->rc = new node();
}
void upd(long long p, pair<long long, long long> val, node *v = rt,
         long long l = 0, long long r = MAXX) {
  v->val = max(val, v->val);
  if (r - l == 1) return;
  long long mid = (l + r) >> 1;
  qwe(v);
  if (p < mid)
    upd(p, val, v->lc, l, mid);
  else
    upd(p, val, v->rc, mid, r);
}
pair<long long, long long> get(long long l, long long r, node *v = rt,
                               long long st = 0, long long ed = MAXX) {
  if (l >= ed || r <= st || (v->val == pair<long long, long long>(0, -1)))
    return {0, -1};
  if (l <= st && ed <= r) return v->val;
  long long mid = (st + ed) >> 1;
  qwe(v);
  return max(get(l, r, v->lc, st, mid), get(l, r, v->rc, mid, ed));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n >> d;
  for (long long i = 0; i < n; i++) cin >> h[i];
  reverse(h, h + n);
  for (long long i = 0; i < n; i++) {
    tie(ans[i], par[i]) = max(get(0, h[i] - d + 1), get(h[i] + d, MAXX));
    ans[i]++;
    upd(h[i], {ans[i], i});
  }
  long long x = max_element(ans, ans + n) - ans;
  cout << ans[x] << '\n';
  while (~par[x]) cout << n - x << ' ', x = par[x];
  cout << n - x;
  return 0;
}
