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 double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const long long INF = (long long)(1e15 + 1);
const int MAXN = 100010;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
struct Node {
  Node *l, *r;
  long long left, right;
  pair<int, int> mx;
  Node() : l(NULL), r(NULL), left(0), right(0), mx() {}
};
void combine(pair<int, int> &a, const pair<int, int> &b) {
  if (a.first < b.first) a = b;
}
void update(Node *v, long long pos, int val, int tpos) {
  if (v->left == pos && v->right == pos) {
    combine(v->mx, make_pair(val, tpos));
    return;
  }
  long long mid = (v->left + v->right) >> 1;
  if (pos <= mid) {
    if (v->l == NULL) v->l = new Node();
    v->l->left = v->left;
    v->l->right = mid;
    update(v->l, pos, val, tpos);
    combine(v->mx, v->l->mx);
  } else {
    if (v->r == NULL) v->r = new Node();
    v->r->left = mid + 1;
    v->r->right = v->right;
    update(v->r, pos, val, tpos);
    combine(v->mx, v->r->mx);
  }
}
pair<int, int> get(Node *v, long long l, long long r) {
  if (v == NULL || v->left > r || l > v->right) return pair<int, int>();
  if (l <= v->left && v->right <= r) return v->mx;
  pair<int, int> res = get(v->l, l, r);
  combine(res, get(v->r, l, r));
  return res;
}
int n, d;
Node *root;
long long a[MAXN];
int dp[MAXN], par[MAXN];
int main() {
  root = new Node();
  root->left = 0;
  root->right = INF;
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  int ans = 0;
  int ansp = 0;
  for (int i = 1; i <= n; i++) {
    long long left = max(0ll, a[i] - d);
    long long right = min(a[i] + d, INF);
    pair<int, int> res = get(root, 1, left);
    combine(res, get(root, right, INF));
    dp[i] = res.first + 1;
    par[i] = res.second;
    update(root, a[i], dp[i], i);
    if (dp[i] > ans) {
      ans = dp[i];
      ansp = i;
    }
  }
  printf("%d\n", ans);
  vector<int> res;
  while (ansp != 0) {
    res.push_back(ansp);
    ansp = par[ansp];
  }
  reverse((res).begin(), (res).end());
  for (int i = 0; i < res.size(); i++) printf("%d ", res[i]);
  return 0;
}
