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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1e9 + 5;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
template <class T>
void chmax(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void chmin(T& a, const T& b) {
  a = min(a, b);
}
int main() {
  int n, d;
  scanf("%d%d", &n, &d);
  vector<long long> nums(n);
  vector<pair<int, int>> dp(n, make_pair(0, 0));
  for (int i = 0; i < n; ++i) scanf("%lld", &nums[i]);
  int len = INT_MIN, index = 0;
  for (int i = 1; i < n; ++i) {
    for (int j = max(0, i - 300); j < i; ++j) {
      if (abs(nums[j] - nums[i]) >= d && dp[j].first + 1 > dp[i].first) {
        dp[i].first = dp[j].first + 1;
        dp[i].second = j;
        if (dp[i].first > len) {
          len = dp[i].first;
          index = i;
        }
      }
    }
  }
  printf("%d\n", len + 1);
  vector<int> res;
  res.push_back(index);
  while (len > 0) {
    index = dp[index].second;
    res.push_back(index);
    len--;
  }
  for (int i = res.size() - 1; i >= 0; --i) {
    printf("%d ", res[i] + 1);
  }
  return 0;
}
