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 maxn = 1e5 + 10;
int pre[maxn];
long long num[maxn];
long long a[maxn];
int cur;
struct node {
  int maxcou;
  int pos;
} tree[maxn << 2];
void build(int rt, int l, int r) {
  if (l == r) {
    tree[rt].maxcou = 0;
    tree[rt].pos = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  tree[rt].maxcou = 0;
  tree[rt].pos = 0;
}
int Rfind(long long x) {
  int l = 1, r = cur;
  while (l < r) {
    int m = (l + r) / 2;
    if (a[m] >= x)
      r = m;
    else
      l = m + 1;
  }
  return l;
}
int Lfind(long long x) {
  int l = 1, r = cur;
  while (l < r) {
    int m = (l + r + 1) / 2;
    if (a[m] <= x)
      l = m;
    else
      r = m - 1;
  }
  return r;
}
void update(int x, int v, int i, int rt, int l, int r) {
  if (l == r) {
    if (v > tree[rt].maxcou) {
      tree[rt].maxcou = v;
      tree[rt].pos = i;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(x, v, i, rt << 1, l, mid);
  else
    update(x, v, i, rt << 1 | 1, mid + 1, r);
  if (tree[rt << 1].maxcou > tree[rt << 1 | 1].maxcou) {
    tree[rt].maxcou = tree[rt << 1].maxcou;
    tree[rt].pos = tree[rt << 1].pos;
  } else {
    tree[rt].maxcou = tree[rt << 1 | 1].maxcou;
    tree[rt].pos = tree[rt << 1 | 1].pos;
  }
}
int maxcount, maxpos;
void query(int L, int R, int rt, int l, int r) {
  if (L > R) return;
  if (L <= l && R >= r) {
    if (tree[rt].maxcou > maxcount) {
      maxcount = tree[rt].maxcou;
      maxpos = tree[rt].pos;
    }
    return;
  }
  int mid = (l + r) >> 1;
  if (L <= mid) query(L, R, rt << 1, l, mid);
  if (R > mid) query(L, R, rt << 1 | 1, mid + 1, r);
}
void prin(int u) {
  if (pre[u] != 0) {
    prin(pre[u]);
    printf("%d ", pre[u]);
  }
  return;
}
int main() {
  int n, k, ids = 1, ans = 1;
  scanf("%d%d", &n, &k);
  memset(pre, 0, sizeof(pre));
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &num[i]);
    a[i] = num[i];
  }
  sort(a + 1, a + n + 1);
  cur = 1;
  for (int i = 2; i <= n; i++)
    if (a[i] != a[cur]) {
      a[++cur] = a[i];
    }
  build(1, 1, cur);
  update(Lfind(num[1]), 1, 1, 1, 1, cur);
  pre[1] = 0;
  for (int i = 2; i <= n; i++) {
    int L = Lfind(num[i] - k);
    int R = Rfind(num[i] + k);
    maxcount = 0, maxpos = 0;
    if (a[L] + k <= num[i]) {
      query(1, L, 1, 1, cur);
    }
    if (num[i] + k <= a[R]) {
      query(R, cur, 1, 1, cur);
    }
    update(Lfind(num[i]), maxcount + 1, i, 1, 1, cur);
    pre[i] = maxpos;
    if (maxcount + 1 > ans) {
      ans = maxcount + 1;
      ids = i;
    }
  }
  printf("%d\n", ans);
  prin(ids);
  printf("%d\n", ids);
  return 0;
}
