#include <bits/stdc++.h>
using namespace std;
class Data {
 public:
  int l, r, pos;
  Data() {}
  Data(int _l, int _r, int _pos) {
    l = _l;
    r = _r;
    pos = _pos;
  }
};
class cmp1 {
 public:
  bool operator()(Data a, Data b) {
    if (a.r == b.r) {
      if (a.l == b.l) {
        return a.pos < b.pos;
      }
      return a.l < b.l;
    }
    return a.r > b.r;
  }
};
set<Data, cmp1> st;
vector<Data> vv[200010];
vector<Data> ri_id[200010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    vv[l].push_back(Data(l, r, i));
    ri_id[r].push_back(Data(l, r, i));
  }
  for (int u = 1; u <= 200000; u++) {
    sort(vv[u].begin(), vv[u].end(), cmp1());
  }
  vector<int> ans;
  for (int u = 1; u <= 200000; u++) {
    for (int i = 0; i <= (int)vv[u].size() - 1; i++) {
      Data data = vv[u][i];
      st.insert(data);
    }
    while (st.size() > k) {
      Data data = *(st.begin());
      st.erase(st.begin());
      ans.push_back(data.pos);
    }
    for (int i = 0; i <= (int)ri_id[u].size() - 1; i++) {
      Data data = ri_id[u][i];
      st.erase(data);
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (int i = 0; i <= (int)ans.size() - 1; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
class Data {
 public:
  int l, r, pos;
  Data() {}
  Data(int _l, int _r, int _pos) {
    l = _l;
    r = _r;
    pos = _pos;
  }
};
class cmp1 {
 public:
  bool operator()(Data a, Data b) {
    if (a.r == b.r) {
      if (a.l == b.l) {
        return a.pos < b.pos;
      }
      return a.l < b.l;
    }
    return a.r > b.r;
  }
};
set<Data, cmp1> st;
vector<Data> vv[200010];
vector<Data> ri_id[200010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    vv[l].push_back(Data(l, r, i));
    ri_id[r].push_back(Data(l, r, i));
  }
  for (int u = 1; u <= 200000; u++) {
    sort(vv[u].begin(), vv[u].end(), cmp1());
  }
  vector<int> ans;
  for (int u = 1; u <= 200000; u++) {
    for (int i = 0; i <= (int)vv[u].size() - 1; i++) {
      Data data = vv[u][i];
      st.insert(data);
    }
    while (st.size() > k) {
      Data data = *(st.begin());
      st.erase(st.begin());
      ans.push_back(data.pos);
    }
    for (int i = 0; i <= (int)ri_id[u].size() - 1; i++) {
      Data data = ri_id[u][i];
      st.erase(data);
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (int i = 0; i <= (int)ans.size() - 1; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

```