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 = 1e5 + 5;
long long arr[N];
vector<pair<long long, int>> vec;
int idx[N];
int tree[4 * N];
int n, d;
int dp[N];
int tillnow[N];
void update(int x, int y, int val, int l = 1, int r = n, int idx = 1) {
  if (l > y || r < x) return;
  if (l >= x && r <= y) {
    tree[idx] = val;
    return;
  }
  int mid = (l + r) >> 1;
  update(x, y, val, l, mid, idx << 1);
  update(x, y, val, mid + 1, r, (idx << 1) + 1);
  tree[idx] = max(tree[idx << 1], tree[(idx << 1) + 1]);
}
int getans(int x, int y, int l = 1, int r = n, int idx = 1) {
  if (l > y || r < x) return 0;
  if (l >= x && r <= y) return tree[idx];
  int mid = (l + r) >> 1;
  return max(getans(x, y, l, mid, idx << 1),
             getans(x, y, mid + 1, r, (idx << 1) + 1));
}
int main() {
  scanf("%d", &n);
  scanf("%d", &d);
  int ans = 0;
  vec.push_back({0, 0});
  for (int i = 1; i <= n; ++i)
    scanf("%lld", &arr[i]), vec.push_back({arr[i], i});
  sort(vec.begin() + 1, vec.end());
  for (int i = 1; i < vec.size(); ++i) idx[vec[i].second] = i;
  for (int i = n; i > 0; --i) {
    long long temp = arr[i];
    long long temp1 = arr[i] - d;
    long long temp2 = arr[i] + d;
    int left, right;
    int l = 1, r = vec.size() - 1, mid;
    while (l < r) {
      mid = l + (r - l + 1) / 2;
      if (vec[mid].first <= temp1)
        l = mid;
      else
        r = mid - 1;
    }
    left = l;
    l = 1, r = vec.size() - 1;
    while (l < r) {
      mid = l + (r - l) / 2;
      if (vec[mid].first >= temp2)
        r = mid;
      else
        l = mid + 1;
    }
    right = l;
    if (vec[left].first <= temp1) dp[i] = max(dp[i], getans(1, left));
    if (vec[right].first >= temp2) dp[i] = max(dp[i], getans(right, n));
    ++dp[i];
    tillnow[i] = dp[i];
    ans = max(ans, dp[i]);
    update(idx[i], idx[i], dp[i]);
  }
  printf("%d\n", ans);
  int prev = -1;
  for (int i = 1; i <= n; ++i) {
    if (tillnow[i] == ans && prev == -1) printf("%d ", i), ans--, prev = i;
    if (tillnow[i] == ans && abs(arr[prev] - arr[i]) >= d)
      printf("%d ", i), ans--, prev = i;
  }
}
