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 = 100010;
const int mod = 1e9 + 7;
int n, d;
long long h[N], tem[N];
int tot, dp[N], pre[N];
int cmp(int x, int y) { return dp[x] < dp[y]; }
struct SegmentTree {
  int v[N << 2];
  void insert(int c, int l, int r, long long height, int i) {
    if (l == r) {
      v[c] = max(v[c], i);
      return;
    }
    int mid = l + r >> 1;
    if (height <= tem[mid])
      insert(c << 1, l, mid, height, i);
    else
      insert(c << 1 | 1, mid + 1, r, height, i);
    v[c] = max(v[c << 1], v[c << 1 | 1], cmp);
  }
  int query(int c, int l, int r, long long height_L, long long height_R) {
    if (l > r) return 0;
    if (height_L < tem[l] && tem[r] < height_R) return 0;
    if (tem[r] <= height_L || height_R <= tem[l]) return v[c];
    int mid = l + r >> 1;
    return max(query(c << 1, l, mid, height_L, height_R),
               query(c << 1 | 1, mid + 1, r, height_L, height_R), cmp);
  }
} st;
int main() {
  scanf("%d %d", &n, &d);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &h[i]);
    tem[i] = h[i];
  }
  sort(tem + 1, tem + n + 1);
  tot = unique(tem + 1, tem + n + 1) - (tem + 1);
  for (int i = 1; i <= n; ++i) {
    pre[i] = st.query(1, 1, tot, h[i] - d, h[i] + d);
    dp[i] = dp[pre[i]] + 1;
    st.insert(1, 1, tot, h[i], i);
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = max(ans, i, cmp);
  printf("%d\n", dp[ans]);
  vector<int> vec;
  while (ans) {
    vec.push_back(ans);
    ans = pre[ans];
  }
  for (int i = (int)vec.size() - 1; i >= 0; --i) printf("%d ", vec[i]);
  return 0;
}
