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>
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline void sub(long long &a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
inline void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
template <typename T>
inline T const &MAX(T const &a, T const &b) {
  return a > b ? a : b;
}
template <typename T>
inline T const &MIN(T const &a, T const &b) {
  return a < b ? a : b;
}
inline long long qp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007, b >>= 1;
  }
  return ans;
}
inline long long qp(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c, b >>= 1;
  }
  return ans;
}
using namespace std;
const unsigned long long ba = 233;
const double eps = 1e-6;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 100000 + 10, maxn = 1000000 + 10, inf = 0x3f3f3f3f;
long long h[N], a[N];
int cnt, pre[N];
struct sgt {
  pair<int, int> ma[N << 2];
  void update(int pos, pair<int, int> v, int l, int r, int rt) {
    if (l == r) {
      ma[rt] = v;
      return;
    }
    int m = (l + r) >> 1;
    if (pos <= m)
      update(pos, v, l, m, rt << 1);
    else
      update(pos, v, m + 1, r, rt << 1 | 1);
    ma[rt] = max(ma[rt << 1], ma[rt << 1 | 1]);
  }
  pair<int, int> query(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) return ma[rt];
    int m = (l + r) >> 1;
    pair<int, int> ans = make_pair(0, 0);
    if (L <= m) ans = max(ans, query(L, R, l, m, rt << 1));
    if (m < R) ans = max(ans, query(L, R, m + 1, r, rt << 1 | 1));
    return ans;
  }
} sg;
int main() {
  int n, d;
  scanf("%d%d", &n, &d);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &h[i]);
    a[cnt++] = h[i];
  }
  sort(a, a + cnt);
  cnt = unique(a, a + cnt) - a;
  for (int i = 1; i <= n; i++) {
    pair<int, int> x = make_pair(0, 0), y;
    int p = upper_bound(a, a + cnt, h[i] - d) - a;
    p--;
    if (p >= 0) x = max(x, sg.query(1, p + 1, 1, cnt, 1));
    p = lower_bound(a, a + cnt, h[i] + d) - a;
    if (p < cnt) x = max(x, sg.query(p + 1, cnt, 1, cnt, 1));
    p = lower_bound(a, a + cnt, h[i]) - a;
    y = sg.query(p + 1, p + 1, 1, cnt, 1);
    if (x.first + 1 > y.first) {
      sg.update(p + 1, make_pair(x.first + 1, i), 1, cnt, 1);
      pre[i] = x.second;
    }
  }
  pair<int, int> x = sg.query(1, cnt, 1, cnt, 1);
  printf("%d\n", x.first);
  stack<int> st;
  for (int now = x.second; now != 0; now = pre[now]) st.push(now);
  while (!st.empty()) printf("%d ", st.top()), st.pop();
  return 0;
}
