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;
class C_ {};
template <typename T>
C_& operator<<(C_& __m, const T& __s) {
  if (!1) cerr << "\E[91m" << __s << "\E[0m";
  return __m;
}
C_ merr;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    merr << "Execution time: " << fixed << setprecision(3)
         << (double)clock() / CLOCKS_PER_SEC << " s.\n";
  }
} __S;
long long A[111111];
long long AA[111111];
int I[111111], II[111111];
int T[444444];
int D[111111];
int P[111111];
int query(int v, int tl, int tr, int l, int r) {
  if (l > r) return 111110;
  if (tl == l && tr == r) {
    return T[v];
  }
  int mm = (tl + tr) / 2;
  int q1 = query(2 * v + 1, tl, mm, l, min(mm, r));
  int q2 = query(2 * v + 2, mm + 1, tr, max(mm + 1, l), r);
  if (D[q1] > D[q2]) return q1;
  return q2;
}
void update(int v, int tl, int tr, int pos, int x) {
  if (tl == tr) {
    T[v] = x;
  } else {
    int mm = (tl + tr) / 2;
    if (pos <= mm) {
      update(2 * v + 1, tl, mm, pos, x);
    } else {
      update(2 * v + 2, mm + 1, tr, pos, x);
    }
    int q1 = T[2 * v + 1], q2 = T[2 * v + 2];
    T[v] = (D[q1] > D[q2] ? q1 : q2);
  }
}
bool comp(int a, int b) { return A[a] < A[b]; }
int main() {
  int n;
  long long d;
  int ans = 0, mi = 111110;
  D[111110] = -1000000000;
  for (int i = 0; i < 111111; i++) {
    I[i] = i;
  }
  for (int i = 0; i < 444444; i++) {
    T[i] = 111110;
  }
  cin >> n >> d;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    AA[i] = A[i];
  }
  sort(I, I + n, comp);
  sort(AA, AA + n);
  for (int i = 0; i < n; i++) {
    II[I[i]] = i;
  }
  for (int i = n - 1; i >= 0; i--) {
    int mx1, mx2;
    int idx1 = n, idx2 = -1, ll, rr;
    ll = 0, rr = n - 1;
    if (AA[n - 1] >= A[i] + d)
      while (ll < rr) {
        int mm = (ll + rr) / 2;
        if (AA[mm] < A[i] + d) {
          ll = mm + 1;
        } else {
          rr = mm;
        }
      }
    if (AA[ll] >= A[i] + d) idx1 = ll;
    mx1 = query(1, 0, n - 1, idx1, n - 1);
    ll = 0, rr = n - 1;
    if (AA[0] <= A[i] - d)
      while (ll < rr) {
        int mm = (ll + rr + 1) / 2;
        if (AA[mm] > A[i] - d) {
          rr = mm - 1;
        } else {
          ll = mm;
        }
      }
    if (AA[rr] <= A[i] - d) idx2 = rr;
    mx2 = query(1, 0, n - 1, 0, idx2);
    if (D[mx2] > D[mx1]) mx1 = mx2;
    D[i] = max(D[mx1], 0) + 1;
    P[i] = mx1;
    update(1, 0, n - 1, II[i], i);
    if (D[i] > ans) {
      ans = D[i];
      mi = i;
    }
  }
  cout << ans << '\n';
  while (mi != 111110) {
    cout << mi + 1 << " ";
    mi = P[mi];
  }
  cout << '\n';
  return 0;
}
