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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int maxn = 100000 + 100;
int n, lim, m, d[maxn], pre[maxn];
long long a[maxn], b[maxn];
int mx[maxn << 2], pos[maxn << 2];
void pushup(int rt) {
  if (mx[(rt << 1)] >= mx[((rt << 1) | 1)])
    mx[rt] = mx[(rt << 1)], pos[rt] = pos[(rt << 1)];
  else
    mx[rt] = mx[((rt << 1) | 1)], pos[rt] = pos[((rt << 1) | 1)];
}
void insert(int x, int p, int v, int l, int r, int rt) {
  if (l == r) {
    if (v > mx[rt]) {
      mx[rt] = v;
      pos[rt] = p;
    }
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    insert(x, p, v, l, mid, (rt << 1));
  else
    insert(x, p, v, mid + 1, r, ((rt << 1) | 1));
  pushup(rt);
}
pair<int, int> query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) return pair<int, int>(mx[rt], pos[rt]);
  int mid = l + r >> 1;
  pair<int, int> ans(0, 0);
  if (L <= mid) ans = max(ans, query(L, R, l, mid, (rt << 1)));
  if (R > mid) ans = max(ans, query(L, R, mid + 1, r, ((rt << 1) | 1)));
  return ans;
}
void print(int x) {
  if (pre[x]) {
    print(pre[x]);
    putchar(' ');
  }
  printf("%d", x);
}
int main() {
  scanf("%d%d", &m, &lim);
  for (int i = 1; i <= m; i++) {
    scanf("%I64d", a + i);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + m);
  n = unique(b + 1, b + 1 + m) - b - 1;
  int len = 0, ans;
  for (int i = 1; i <= m; i++) {
    int L = upper_bound(b + 1, b + 1 + n, a[i] - lim) - b;
    int R = lower_bound(b + 1, b + 1 + n, a[i] + lim) - b;
    d[i] = -1;
    if (L > 1) {
      pair<int, int> t = query(1, L - 1, 1, n, 1);
      if (t.first + 1 > d[i]) d[i] = t.first + 1, pre[i] = t.second;
    }
    if (R <= n) {
      pair<int, int> t = query(R, n, 1, n, 1);
      if (t.first + 1 > d[i]) d[i] = t.first + 1, pre[i] = t.second;
    }
    int k = lower_bound(b + 1, b + 1 + n, a[i]) - b;
    insert(k, i, d[i], 1, n, 1);
    if (d[i] > len) len = d[ans = i];
  }
  printf("%d\n", len);
  print(ans);
  puts("");
  return 0;
}
