#include <bits/stdc++.h>
using namespace std;
int n, k, tree[800005], lazy[800005], maxi;
vector<pair<int, int> > a[200005];
vector<int> ans;
void pushdown(int ind, int l, int r) {
  if (l == r)
    tree[ind] += lazy[ind];
  else {
    lazy[2 * ind] += lazy[ind];
    lazy[2 * ind + 1] += lazy[ind];
  }
  lazy[ind] = 0;
}
void update(int ind, int l, int r, int ll, int rr, int val) {
  pushdown(ind, l, r);
  if (l > rr || r < ll) return;
  if (ll <= l && r <= rr) {
    lazy[ind] += val;
    pushdown(ind, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * ind, l, mid, ll, rr, val);
  update(2 * ind + 1, mid + 1, r, ll, rr, val);
}
int query(int ind, int l, int r, int pos) {
  pushdown(ind, l, r);
  if (l > pos || r < pos) return 0;
  if (l == r && l == pos) return tree[ind];
  int mid = (l + r) / 2;
  return query(2 * ind, l, mid, pos) + query(2 * ind + 1, mid + 1, r, pos);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int l, r;
    cin >> l >> r;
    maxi = max(maxi, r);
    a[l].push_back({r, i});
    update(1, 1, 200005, l, r, 1);
  }
  set<pair<int, int> > s;
  for (int i = 1; i <= maxi; ++i) {
    for (auto j : a[i]) s.insert(j);
    int t = query(1, 1, 200005, i);
    while (t > k) {
      auto tmp = *s.rbegin();
      s.erase(tmp);
      update(1, 1, 200005, i, tmp.first, -1);
      ans.push_back(tmp.second);
      t--;
    }
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << " ";
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
4 6 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 4 5 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, k, tree[800005], lazy[800005], maxi;
vector<pair<int, int> > a[200005];
vector<int> ans;
void pushdown(int ind, int l, int r) {
  if (l == r)
    tree[ind] += lazy[ind];
  else {
    lazy[2 * ind] += lazy[ind];
    lazy[2 * ind + 1] += lazy[ind];
  }
  lazy[ind] = 0;
}
void update(int ind, int l, int r, int ll, int rr, int val) {
  pushdown(ind, l, r);
  if (l > rr || r < ll) return;
  if (ll <= l && r <= rr) {
    lazy[ind] += val;
    pushdown(ind, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * ind, l, mid, ll, rr, val);
  update(2 * ind + 1, mid + 1, r, ll, rr, val);
}
int query(int ind, int l, int r, int pos) {
  pushdown(ind, l, r);
  if (l > pos || r < pos) return 0;
  if (l == r && l == pos) return tree[ind];
  int mid = (l + r) / 2;
  return query(2 * ind, l, mid, pos) + query(2 * ind + 1, mid + 1, r, pos);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int l, r;
    cin >> l >> r;
    maxi = max(maxi, r);
    a[l].push_back({r, i});
    update(1, 1, 200005, l, r, 1);
  }
  set<pair<int, int> > s;
  for (int i = 1; i <= maxi; ++i) {
    for (auto j : a[i]) s.insert(j);
    int t = query(1, 1, 200005, i);
    while (t > k) {
      auto tmp = *s.rbegin();
      s.erase(tmp);
      update(1, 1, 200005, i, tmp.first, -1);
      ans.push_back(tmp.second);
      t--;
    }
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << " ";
  return 0;
}

```