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 int INF = 2000000000;
const int MOD = 10007;
long long DATA[100007], ODATA[100007], N, D;
long long DP[100007];
map<long long, int> MPP;
int M, tot;
vector<int> RES;
struct BIT {
  vector<int> T;
  BIT(int N) { T.resize(N + 7, 0); }
  int lastone(int x) { return x & (x ^ (x - 1)); }
  void addsum(int pos, int val) {
    for (int i = pos + 1; i < T.size(); i += lastone(i)) T[i] = max(T[i], val);
  }
  int getsum(int pos) {
    int ret = 0;
    for (int i = pos + 1; i > 0; i -= lastone(i)) ret = max(ret, T[i]);
    return ret;
  }
};
int main(int argc, const char **argv) {
  int pos, ls, rs;
  long long resl, resr;
  long long ans;
  while (cin >> tot >> D) {
    MPP.clear();
    RES.clear();
    for (int i = 1; i <= tot; ++i) {
      cin >> DATA[i];
      ODATA[i] = DATA[i];
      MPP[DATA[i]] = 0;
      MPP[(DATA[i] + D)] = 0;
      MPP[(DATA[i] - D)] = 0;
    }
    M = 0;
    pos = -1;
    ans = 0;
    for (map<long long, int>::iterator itt = MPP.begin(); itt != MPP.end();
         itt++) {
      itt->second = M++;
    }
    M++;
    memset(DP, 0, sizeof(DP));
    BIT F1(M), F2(M);
    for (int i = 1; i <= tot; ++i) {
      ls = MPP[(DATA[i] - D)];
      rs = M - MPP[(DATA[i] + D)];
      DATA[i] = MPP[DATA[i]];
      resl = F1.getsum(ls);
      resr = F2.getsum(rs);
      DP[i] = max(resl, resr) + 1;
      ans = max(ans, DP[i]);
      if (ans == DP[i]) pos = i;
      F1.addsum(DATA[i], DP[i]);
      F2.addsum(M - DATA[i], DP[i]);
    }
    RES.push_back(pos);
    for (int jaga = pos - 1; jaga >= 1; jaga--) {
      if (abs(ODATA[jaga] - ODATA[pos]) >= D && DP[jaga] == (ans - 1)) {
        RES.push_back(jaga);
        ans--;
        pos = jaga;
      }
    }
    reverse(RES.begin(), RES.end());
    ;
    cout << ((int)RES.size()) << "\n";
    for (int i = 0; i <= (((int)RES.size()) - 1); ++i) cout << RES[i] << " ";
    cout << "\n";
  }
  return 0;
}
