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 h[100010];
long long tmp[100010];
int dp[100010];
vector<int> path;
int tree[100010 << 2];
inline void push_up(int rt) {
  tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
}
void update(int x, int val, int L, int R, int rt) {
  if (L == R) {
    tree[rt] = val;
  } else {
    int mid = (L + R) >> 1;
    if (x <= mid) {
      update(x, val, L, mid, rt << 1);
    } else {
      update(x, val, mid + 1, R, rt << 1 | 1);
    }
    push_up(rt);
  }
}
int query_max(int x, int y, int L, int R, int rt) {
  if (x <= L && R <= y) {
    return tree[rt];
  } else {
    int mid = (L + R) >> 1;
    int ans = 0;
    if (x <= mid) {
      ans = max(ans, query_max(x, y, L, mid, rt << 1));
    }
    if (y > mid) {
      ans = max(ans, query_max(x, y, mid + 1, R, rt << 1 | 1));
    }
    return ans;
  }
}
int main() {
  int n, d;
  while (cin >> n >> d) {
    for (int i = 0; i < n; i++) {
      cin >> h[i];
      tmp[i] = h[i];
    }
    sort(tmp, tmp + n);
    memset(tree, 0, sizeof(tree));
    for (int i = 0; i < n; i++) {
      dp[i] = 0;
      int x0 = lower_bound(tmp, tmp + n, h[i] + d) - tmp;
      if (x0 <= n - 1) {
        dp[i] = max(dp[i], query_max(x0, n - 1, 0, n - 1, 1));
      }
      int x1 = lower_bound(tmp, tmp + n, h[i] - d + 1) - tmp - 1;
      if (x1 >= 0) {
        dp[i] = max(dp[i], query_max(0, x1, 0, n - 1, 1));
      }
      int x2 = lower_bound(tmp, tmp + n, h[i]) - tmp;
      dp[i]++;
      update(x2, dp[i], 0, n - 1, 1);
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      ans = max(ans, dp[i]);
    }
    cout << ans << endl;
    for (int i = 0; i < n; i++) {
      if (dp[i] == ans) {
        path.clear();
        path.push_back(i + 1);
        long long val = h[i];
        ans--;
        for (i--; i >= 0; i--) {
          if (dp[i] == ans && abs(val - h[i]) >= d) {
            path.push_back(i + 1);
            ans--;
            val = h[i];
          }
        }
        break;
      }
    }
    for (int i = path.size() - 1; i >= 0; i--) {
      cout << path[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
