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 GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("unroll-loops")
using namespace std;
void my_file_in_out() {
  freopen("inp.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
inline void OK(string t = "") { cerr << "OK " << t << '\n'; }
inline long double time() { return clock() / (long double)CLOCKS_PER_SEC; }
inline void stop(string t = "PROCESS FINISHED") {
  cerr << t << '\n';
  exit(0);
}
long long bp(long long c, long long st) {
  if (!st) return 1;
  if (st & 1) return c * bp(c, st - 1);
  return bp(c * c, st >> 1);
}
namespace VECTOR_OUT {
template <class T>
ostream &operator<<(ostream &out, vector<T> &a) {
  for (auto &elem : a) out << elem << ' ';
  out << endl;
  return out;
}
template <class T>
ostream &operator<<(ostream &out, vector<vector<T>> &a) {
  for (auto &elem : a) out << elem;
  return out;
}
}  // namespace VECTOR_OUT
using namespace VECTOR_OUT;
const int MAXN = 7e6;
const long long INF = 1e15 + 1;
struct node {
  int lc, rc, ps, mx;
  node() {
    lc = rc = 0;
    mx = 0;
    ps = -1;
  }
};
node tree[MAXN];
int SIZE = 1;
void merge(int &T, int &T1, int &T2) {
  if (T == 0) return;
  if (tree[T1].mx > tree[T2].mx) {
    tree[T].mx = tree[T1].mx;
    tree[T].ps = tree[T1].ps;
  } else {
    tree[T].mx = tree[T2].mx;
    tree[T].ps = tree[T2].ps;
  }
}
void update(int T, long long lg, long long rg, long long p, int x, int c) {
  if (lg == rg) {
    if (tree[T].mx < c) {
      tree[T].ps = x;
      tree[T].mx = c;
    }
    return;
  }
  long long m = (lg + rg) >> 1;
  if (tree[T].lc == 0) {
    tree[T].lc = SIZE++;
    tree[T].rc = SIZE++;
  }
  if (p <= m)
    update(tree[T].lc, lg, m, p, x, c);
  else
    update(tree[T].rc, m + 1, rg, p, x, c);
  merge(T, tree[T].lc, tree[T].rc);
}
pair<int, int> query(int T, long long lg, long long rg, long long l,
                     long long r) {
  if (lg > r || rg < l) return {0, -1};
  if (lg >= l && rg <= r) return {tree[T].mx, tree[T].ps};
  if (tree[T].lc == 0) return {0, -1};
  long long m = (lg + rg) >> 1;
  auto a1 = query(tree[T].lc, lg, m, l, r);
  auto a2 = query(tree[T].rc, m + 1, rg, l, r);
  return max(a1, a2);
}
signed main() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long lg = 0;
  long long rg = INF;
  int T = SIZE++;
  int n, d;
  cin >> n >> d;
  vector<long long> h(n);
  for (int i = 0; i < n; ++i) {
    cin >> h[i];
    h[i] += (d + 1);
  }
  vector<int> dp(n, 0);
  vector<int> pr(n, -1);
  for (int i = 0; i < n; ++i) {
    auto res1 = query(T, lg, rg, 0, h[i] - d);
    auto res2 = query(T, lg, rg, h[i] + d, INF);
    if (res1 > res2) {
      dp[i] = res1.first + 1;
      pr[i] = res1.second;
    } else {
      dp[i] = res2.first + 1;
      pr[i] = res2.second;
    }
    update(T, lg, rg, h[i], i, dp[i]);
  }
  int id = 0;
  for (int i = 0; i < n; ++i)
    if (dp[i] > dp[id]) id = i;
  vector<int> ans;
  while (id != -1) {
    ans.push_back(id + 1);
    id = pr[id];
  }
  reverse(ans.begin(), ans.end());
  cout << ((int)ans.size()) << '\n' << ans;
  return 0;
}
