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 = 100005;
const int mod = 1000000007;
int n;
long long d, a[maxn];
vector<long long> vis;
int f[maxn], pnt[maxn];
struct ST {
  int l, r, val, idx;
} st[maxn * 4];
void build(int id, int l, int r) {
  st[id].l = l, st[id].r = r, st[id].val = -1, st[id].idx = -1;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(id << 1, l, m), build(id << 1 | 1, m + 1, r);
}
void upd(int id, int pos, int val, int idx) {
  if (st[id].l == st[id].r) {
    if (st[id].val < val) {
      st[id].val = val, st[id].idx = idx;
    }
    return;
  }
  int m = (st[id].l + st[id].r) >> 1;
  if (pos <= m)
    upd(id << 1, pos, val, idx);
  else
    upd(id << 1 | 1, pos, val, idx);
  if (st[id << 1].val < st[id << 1 | 1].val) {
    st[id].val = st[id << 1 | 1].val, st[id].idx = st[id << 1 | 1].idx;
  } else {
    st[id].val = st[id << 1].val, st[id].idx = st[id << 1].idx;
  }
}
void init() {
  scanf("%d%lld", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    vis.push_back(a[i]);
  }
  sort(vis.begin(), vis.end());
  vis.erase(unique(vis.begin(), vis.end()), vis.end());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(vis.begin(), vis.end(), a[i]) - vis.begin() + 1;
  build(1, 1, vis.size());
}
int val, idx;
void query(int id, int l, int r) {
  if (st[id].l == l && st[id].r == r) {
    if (st[id].val > val) {
      val = st[id].val, idx = st[id].idx;
    }
    return;
  }
  int m = (st[id].l + st[id].r) >> 1;
  if (r <= m)
    query(id << 1, l, r);
  else if (l > m)
    query(id << 1 | 1, l, r);
  else
    query(id << 1, l, m), query(id << 1 | 1, m + 1, r);
}
int bs_1(long long x) {
  int l = 0, r = vis.size() - 1, ret = -1;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (vis[m] <= x) {
      ret = m;
      l = m + 1;
    } else
      r = m - 1;
  }
  return ret;
}
int bs_2(long long x) {
  int l = 0, r = vis.size() - 1, ret = -1;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (vis[m] >= x) {
      ret = m;
      r = m - 1;
    } else
      l = m + 1;
  }
  return ret;
}
vector<int> ans;
int main() {
  init();
  for (int i = 1; i <= n; i++) {
    f[i] = 1;
    long long tt = vis[a[i] - 1];
    val = -1, idx = -1;
    int r = bs_1(tt - d);
    if (r != -1) query(1, 1, r + 1);
    int l = bs_2(tt + d);
    if (l != -1) query(1, l + 1, vis.size());
    if (f[i] < val + 1) {
      f[i] = val + 1;
      pnt[i] = idx;
    }
    upd(1, (int)a[i], f[i], i);
  }
  int mx = 0, idx;
  for (int i = 1; i <= n; i++)
    if (mx < f[i]) {
      mx = f[i], idx = i;
    }
  printf("%d\n", mx);
  while (idx) {
    ans.push_back(idx);
    idx = pnt[idx];
  }
  for (int i = mx - 1; i >= 0; i--) printf("%d%c", ans[i], i == 0 ? '\n' : ' ');
  return 0;
}
