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 N = 100002, oo = 1 << 29;
int n, d;
long long arr[N];
vector<long long> v;
int dp[N];
int seg[4 * N];
int idx, value;
int from[N];
void build(int node, int i, int j) {
  if (i > j) return;
  if (i == j) {
    seg[node] = i;
    return;
  }
  build(node * 2, i, (i + j) / 2);
  build(node * 2 + 1, (i + j) / 2 + 1, j);
  if (dp[seg[node * 2]] > dp[seg[node * 2 + 1]])
    seg[node] = seg[node * 2];
  else
    seg[node] = seg[node * 2 + 1];
}
void update(int node, int i, int j) {
  if (i > idx || j < idx) return;
  if (i == idx && j == idx) return;
  update(node * 2, i, (i + j) / 2);
  update(node * 2 + 1, (i + j) / 2 + 1, j);
  if (dp[seg[node * 2]] > dp[seg[node * 2 + 1]])
    seg[node] = seg[node * 2];
  else
    seg[node] = seg[node * 2 + 1];
}
int F, T;
int get(int node, int i, int j) {
  if (i > T || j < F || i > j) return -1;
  if (i >= F && j <= T) return seg[node];
  int L = get(node * 2, i, (i + j) / 2);
  int R = get(node * 2 + 1, (i + j) / 2 + 1, j);
  if (L == -1) return R;
  if (R == -1) return L;
  if (dp[L] > dp[R])
    return L;
  else
    return R;
}
int last[N];
void road(int x) {
  int z;
  if (from[x] == -1) {
    cout << x + 1 << " " << endl;
    return;
  }
  road(from[x]);
  cout << x + 1 << " " << endl;
}
int main() {
  scanf("%d %d", &n, &d);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", arr + i);
    v.push_back(arr[i]);
  }
  memset(from, -1, sizeof(from));
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  build(1, 0, (int)v.size() - 1);
  for (int i = 0; i < n; ++i) {
    F = 0;
    T = lower_bound(v.begin(), v.end(), arr[i] - d) - v.begin();
    if (v[T] != (arr[i] - d)) --T;
    int idx1 = get(1, 0, v.size() - 1);
    int val1 = (idx1 != -1) ? dp[idx1] : 0;
    F = lower_bound(v.begin(), v.end(), arr[i] + d) - v.begin();
    T = (int)v.size() - 1;
    int idx2, val2;
    if (F <= T) {
      idx2 = get(1, 0, v.size() - 1);
      val2 = (idx2 != -1) ? dp[idx2] : 0;
    }
    if (F > T) val2 = 0;
    idx = lower_bound(v.begin(), v.end(), arr[i]) - v.begin();
    if (val1 >= val2) {
      from[i] = (val1 != 0) ? last[idx1] : -1;
    } else {
      from[i] = (val2 != 0) ? last[idx2] : -1;
    }
    dp[idx] = 1 + max(val1, val2);
    update(1, 0, v.size() - 1);
    last[idx] = i;
  }
  int mx = -1;
  for (int i = 0; i < N; ++i) mx = max(mx, dp[i]);
  cout << mx << endl;
  for (int i = n - 1; i >= 0; --i)
    if (dp[lower_bound(v.begin(), v.end(), arr[i]) - v.begin()] == mx) {
      road(i);
      break;
    }
  return 0;
}
