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 n, k, l, r, sz, i, len, q, p, mid, anspos, ans[1000001];
long long mxl, mxr;
long long pos[1000001], t[1000001], pr[100001];
struct Pair {
  long long pos;
  long long x;
  long long d;
};
Pair a[100001], b[100001];
bool cmp(Pair p1, Pair p2) { return p1.x < p2.x; }
bool cmd(Pair p1, Pair p2) { return p1.pos < p2.pos; }
long long binl(long long L, long long R, long long height) {
  long long otv = -1;
  while (L <= R) {
    mid = L + R;
    mid /= 2;
    if (b[mid].x > height)
      R = mid - 1;
    else
      otv = mid, L = mid + 1;
  }
  return otv;
}
long long binr(long long L, long long R, long long height) {
  long long otv = -1;
  while (L <= R) {
    mid = L + R;
    mid /= 2;
    if (b[mid].x < height)
      L = mid + 1;
    else
      otv = mid, R = mid - 1;
  }
  return otv;
}
long long tree(long long L, long long R) {
  long long nn = 0;
  while (L <= R) {
    if (L % 2 == 1) nn = max(nn, t[L]);
    if (R % 2 == 0) nn = max(nn, t[R]);
    L = (L + 1) / 2;
    R = (R - 1) / 2;
  }
  return nn;
}
void update(long long x) {
  x /= 2;
  while (x) {
    t[x] = max(t[x * 2], t[x * 2 + 1]);
    x /= 2;
  }
}
int main() {
  cin >> n >> k;
  sz = 1;
  while (sz < n) sz *= 2;
  for (i = 1; i <= n; i++) {
    cin >> a[i].x;
    a[i].pos = i;
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) pos[b[i].pos] = i, pr[i] = b[i].pos;
  for (int i = 1; i <= n; i++) {
    int l = binl(1, pos[i] - 1, a[i].x - k);
    int r = binr(pos[i] + 1, n, a[i].x + k);
    int mxl = 0, mxr = 0;
    if (l > 0) mxl = tree(sz, l + sz - 1);
    if (r > 0) mxr = tree(r + sz - 1, n + sz - 1);
    t[pos[i] + sz - 1] = max(mxl, mxr) + 1;
    update(pos[i] + sz - 1);
    a[pr[pos[i]]].d = t[pos[i] + sz - 1];
  }
  sort(b + 1, b + n + 1, cmd);
  cout << t[1] << endl;
  for (int i = 1; i <= n; i++) {
    if (a[i].d == t[1]) {
      p = i;
      break;
    }
  }
  ans[++q] = p;
  for (int j = p; j >= 1; j--) {
    if (a[j].d == a[p].d - 1 && abs(a[j].x - a[p].x) >= k) {
      ans[++q] = j;
      p = j;
    }
  }
  for (int i = q; i >= 1; i--) cout << ans[i] << " ";
  return 0;
}
