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;
int A[100005], P[100005], N[100005], T[262150];
long long H[100005], B[100005];
void build_segment(int v, int l, int r) {
  if (l == r)
    T[v] = l;
  else {
    build_segment(((v) << 1), l, (l + r) / 2),
        build_segment((((v) << 1) | 1), (l + r) / 2 + 1, r);
    T[v] = (A[T[((v) << 1)]] >= A[T[(((v) << 1) | 1)]] ? T[((v) << 1)]
                                                       : T[(((v) << 1) | 1)]);
  }
}
void build_ST(int n) { build_segment(1, 0, n - 1); }
int query(int v, int l, int r, int i, int j) {
  if (r < i || l > j) return -1;
  if (l >= i && r <= j) return T[v];
  int p1 = query(((v) << 1), l, (l + r) / 2, i, j),
      p2 = query((((v) << 1) | 1), (l + r) / 2 + 1, r, i, j);
  if (p1 == -1) return p2;
  if (p2 == -1) return p1;
  return A[p1] >= A[p2] ? p1 : p2;
}
void update(int v, int l, int r, int idx, int val) {
  if (l == r)
    A[l] = val;
  else {
    if (idx <= (l + r) / 2)
      update(((v) << 1), l, (l + r) / 2, idx, val);
    else
      update((((v) << 1) | 1), (l + r) / 2 + 1, r, idx, val);
    T[v] = (A[T[((v) << 1)]] >= A[T[(((v) << 1) | 1)]] ? T[((v) << 1)]
                                                       : T[(((v) << 1) | 1)]);
  }
}
void solve(int n, int d) {
  int c = 0, maxLen = 0, start, p, p1, p2, p3, p4, i, j;
  map<long long, int> M;
  memcpy(B, H, n * sizeof(long long));
  sort(B, B + n);
  for (i = 0; i < n; i = j) {
    for (j = i + 1; j < n && B[j] == B[i]; ++j)
      ;
    M[B[i]] = ++c;
  }
  build_ST(c + 1);
  for (i = n - 1; i >= 0; --i) {
    p1 = lower_bound(B, B + n, H[i] + d) - B,
    p2 = upper_bound(B, B + n, H[i] - d) - B - 1;
    p3 = (p1 == n ? 0 : query(1, 0, c, M[B[p1]], c)),
    p4 = (p2 < 0 ? 0 : query(1, 0, c, 1, M[B[p2]]));
    p = (A[p3] >= A[p4] ? p3 : p4);
    if (maxLen < 1 + A[p]) maxLen = 1 + A[p], start = i;
    update(1, 0, c, M[H[i]], 1 + A[p]), N[i] = P[p];
    P[M[H[i]]] = i;
  }
  printf("%d\n", maxLen);
  p = start;
  do {
    printf("%d ", p + 1);
    p = N[p];
  } while (p);
}
int main() {
  int n, d, i;
  scanf("%d %d", &n, &d);
  for (i = 0; i < n; ++i) scanf("%I64d", H + i);
  solve(n, d);
  return 0;
}
