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 long long N = 100100;
const long long Inf = 1e16;
long long n, d;
struct TREE {
  TREE* Left;
  TREE* Right;
  long long Value;
  long long Index;
  TREE() : Left(NULL), Right(NULL), Value(0), Index(0) {}
  TREE(long long V, long long I)
      : Left(NULL), Right(NULL), Value(V), Index(I) {}
  TREE(TREE* L, TREE* R, long long V, long long I)
      : Left(L), Right(R), Value(V), Index(I) {}
};
TREE* Root = new TREE(NULL, NULL, 0, 0);
void max_self(TREE* p1, TREE* p2) {
  if (p1->Value <= p2->Value) p1->Value = p2->Value, p1->Index = p2->Index;
}
void Merge(TREE* p) {
  if (p->Left != NULL) max_self(p, p->Left);
  if (p->Right != NULL) max_self(p, p->Right);
}
void upd(long long i, long long v, long long id, TREE*& p = Root,
         long long L = 1, long long R = Inf) {
  if (i < L || R < i) return;
  if (p == NULL) p = new TREE();
  assert(p != NULL);
  if (L == R) {
    TREE tmp = TREE(v, id);
    max_self(p, &tmp);
    return;
  }
  long long Mid = (L + R) / 2;
  upd(i, v, id, p->Left, L, Mid);
  upd(i, v, id, p->Right, Mid + 1, R);
  Merge(p);
}
pair<long long, long long> Query(long long i, long long j, TREE* p = Root,
                                 long long L = 1, long long R = Inf) {
  if (p == NULL) return pair<long long, long long>(0, n + 1);
  if (j < L || R < i) return pair<long long, long long>(0, n + 1);
  if (i <= L && R <= j) return pair<long long, long long>(p->Value, p->Index);
  long long Mid = (L + R) / 2;
  return max(Query(i, j, p->Left, L, Mid), Query(i, j, p->Right, Mid + 1, R));
}
void Path(pair<long long, long long> p);
long long h[N];
pair<long long, long long> Dp[N];
int main() {
  cin >> n >> d;
  for (long long i = 1; i <= n; i++) scanf("%I64d", h + i);
  pair<long long, long long> Max = pair<long long, long long>(0, n + 1);
  for (long long i = n; i >= 1; i--) {
    Dp[i] = max(
        Query(h[i] + d, Inf),
        h[i] > d ? Query(1, h[i] - d) : pair<long long, long long>(0, n + 1));
    Dp[i].first++;
    Max = max(Max, pair<long long, long long>(Dp[i].first, i));
    upd(h[i], Dp[i].first, i);
  }
  cout << Max.first << endl;
  Path(Max);
}
void Path(pair<long long, long long> p) {
  if (p.second == n + 1) return;
  return printf("%I64d ", p.second), Path(Dp[p.second]);
}
