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;
long long v[262144], w[262144], uniq[262144];
long long k, n, d, pos[262144], D[262144];
long long arb[262144], up[262144];
int getleft(long long x) {
  int l = 0, r = k + 1;
  int p = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (x >= uniq[mid]) {
      l = mid + 1;
      p = mid;
    }
    if (x < uniq[mid]) {
      r = mid - 1;
    }
  }
  return p;
}
int getright(long long x) {
  int l = 0, r = k + 1;
  int p = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (x > uniq[mid]) {
      l = mid + 1;
    }
    if (x <= uniq[mid]) {
      p = mid;
      r = mid - 1;
    }
  }
  return p;
}
void update(int n, int l, int r, int poz, int val) {
  if (l == r) {
    arb[n] = val;
    pos[n] = poz;
    return;
  }
  int mid = (l + r) / 2;
  if (poz <= mid) update(2 * n, l, mid, poz, val);
  if (mid < poz) update(2 * n + 1, mid + 1, r, poz, val);
  if (arb[2 * n] > arb[2 * n + 1]) {
    pos[n] = pos[2 * n];
    arb[n] = arb[2 * n];
  } else {
    pos[n] = pos[2 * n + 1];
    arb[n] = arb[2 * n + 1];
  }
}
int query_result, query_max;
void query(int n, int l, int r, int st, int end) {
  if (st <= l && r <= end) {
    if (arb[n] > query_max) {
      query_max = arb[n];
      query_result = pos[n];
    }
    return;
  }
  int mid = (l + r) / 2;
  if (st <= mid) {
    query(2 * n, l, mid, st, end);
  }
  if (mid < end) {
    query(2 * n + 1, mid + 1, r, st, end);
  }
}
inline long long myabs(long long x) { return (x > 0) ? x : -x; }
int main() {
  cin.sync_with_stdio(0);
  cin >> n >> d;
  for (int i = 1; i <= n; i++) cin >> v[i], w[i] = v[i];
  sort(w + 1, w + n + 1);
  for (int i = 1; i <= n; i++) {
    if (w[i] != uniq[k]) {
      uniq[++k] = w[i];
    }
  }
  for (int i = 1; i <= n; i++) {
    v[i] = lower_bound(uniq + 1, uniq + k + 1, v[i]) - uniq;
  }
  uniq[k + 1] = 0x3f3f3f3f3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    int l = getleft(uniq[v[i]] - d);
    int r = getright(uniq[v[i]] + d);
    query_max = -1;
    if (1 <= l) query(1, 1, k, 1, l);
    if (r <= k) query(1, 1, k, r, k);
    D[i] = query_max + 1;
    update(1, 1, k, v[i], D[i]);
  }
  int bst = 0;
  for (int i = 1; i <= n; i++) {
    if (D[bst] < D[i]) bst = i;
  }
  vector<int> sol;
  sol.push_back(bst);
  for (int i = bst - 1; i >= 1; i--) {
    if (myabs(uniq[v[i]] - uniq[v[bst]]) >= d && D[bst] == D[i] + 1) {
      sol.push_back(i);
      bst = i;
    }
  }
  cout << sol.size() << endl;
  for (int i = sol.size() - 1; i >= 0; --i) cout << sol[i] << " ";
  return 0;
}
