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 dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long int lxt() {
  long long int aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
long long int ar[200010], br[200010];
struct info {
  long long int sum;
  long long int mx;
} tree[200010 * 4];
void update(int node, int low, int hi, int i, int value) {
  if (low == hi) {
    tree[node].sum = max(tree[node].sum, value * 1LL);
    return;
  }
  int mid = (low + hi) / 2;
  int left = 2 * node;
  int right = left + 1;
  if (i <= mid)
    update(left, low, mid, i, value);
  else
    update(right, mid + 1, hi, i, value);
  tree[node].sum = max(tree[left].sum, tree[right].sum);
}
int query(int node, int low, int hi, int i, int j) {
  if (i > hi || j < low) return 0;
  if (low >= i && hi <= j) return tree[node].sum;
  int mid = (low + hi) / 2;
  int left = 2 * node;
  int right = left + 1;
  long long int x = query(left, low, mid, i, j);
  long long int y = query(right, mid + 1, hi, i, j);
  return max(x, y);
}
map<long long int, int> mp;
int dp[200010];
int main() {
  int n = nxt();
  long long int d = lxt();
  for (int i = 0; i < n; i++) {
    ar[i] = lxt();
    br[i] = ar[i];
    mp[ar[i]] = 1;
  }
  sort(br, br + n);
  int rnk = 1;
  for (auto it : mp) {
    mp[it.first] = rnk++;
  }
  int ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long int val = ar[i];
    int l = lower_bound(br, br + n, val + d) - br;
    int mx = 0;
    if (l < n) {
      mx = max(mx, query(1, 1, rnk, mp[br[l]], rnk));
    }
    l = lower_bound(br, br + n, val - d) - br;
    if (val == 7)
      ;
    if (abs(br[l] - val) < d) {
      l--;
    }
    if (l >= 0) mx = max(mx, query(1, 1, rnk, 1, mp[br[l]]));
    dp[i] = mx + 1;
    update(1, 1, rnk, mp[val], dp[i]);
    ;
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
  long long int last = -3e17;
  for (int i = 0; i < n; i++) {
    if (dp[i] == ans) {
      if (abs(last - ar[i]) >= d) {
        printf("%d ", i + 1);
        last = ar[i];
        ans--;
      }
    }
  }
  return 0;
}
