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;
long long INF = (1e15) + 1;
int n, dp[100004], par[100004];
long long d, arr[100004];
struct Node {
  Node *leftNode, *rightNode;
  long long leftBound, rightBound;
  pair<int, int> mx;
  Node() {
    leftNode = rightNode = NULL;
    leftBound = rightBound = 0LL;
    mx = {0, 0};
  }
};
void update(Node *par, long long pos, int dpMax, int ind) {
  if (par->leftBound == pos && par->rightBound == pos) {
    if ((par->mx).first < dpMax) {
      par->mx = {dpMax, ind};
    }
    return;
  }
  long long mid = (par->leftBound + par->rightBound) >> 1;
  if (pos <= mid) {
    if (par->leftNode == NULL) par->leftNode = new Node();
    par->leftNode->leftBound = par->leftBound;
    par->leftNode->rightBound = mid;
    update(par->leftNode, pos, dpMax, ind);
    if ((par->leftNode->mx).first > (par->mx).first) {
      par->mx = par->leftNode->mx;
    }
  } else {
    if (par->rightNode == NULL) par->rightNode = new Node();
    par->rightNode->leftBound = mid + 1;
    par->rightNode->rightBound = par->rightBound;
    update(par->rightNode, pos, dpMax, ind);
    if ((par->rightNode->mx).first > (par->mx).first) {
      par->mx = par->rightNode->mx;
    }
  }
}
pair<int, int> query(Node *par, long long l, long long r) {
  if (par == NULL || par->leftBound > r || par->rightBound < l) {
    return {0, 0};
  }
  if (par->leftBound >= l && par->rightBound <= r) {
    return par->mx;
  }
  long long mid = (par->leftBound + par->rightBound) >> 1;
  pair<int, int> p1 = query(par->leftNode, l, r);
  pair<int, int> p2 = query(par->rightNode, l, r);
  if (p1.first > p2.first)
    return p1;
  else
    return p2;
}
int main() {
  scanf("%d%lld", &n, &d);
  for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  Node *root = new Node();
  root->leftBound = 0LL;
  root->rightBound = INF;
  int ans = 0, curInd = 0;
  for (int i = 1; i <= n; i++) {
    long long l = max(0LL, arr[i] - d);
    long long r = min(INF, arr[i] + d);
    pair<int, int> p1 = query(root, 1LL, l);
    pair<int, int> p2 = query(root, r, INF);
    pair<int, int> p = p1;
    if (p.first < p2.first) p = p2;
    dp[i] = p.first + 1;
    par[i] = p.second;
    update(root, arr[i], dp[i], i);
    if (dp[i] > ans) {
      ans = dp[i];
      curInd = i;
    }
  }
  printf("%d\n", ans);
  vector<int> inds;
  while (curInd != 0) {
    inds.push_back(curInd);
    curInd = par[curInd];
  }
  reverse(inds.begin(), inds.end());
  for (int i = 0; i <= inds.size() - 1; i++) {
    printf("%d ", inds[i]);
  }
  return 0;
}
