#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmd(long long a, long long b) {
  a = a % md;
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) {
      ret = ret * b % md;
    }
    b = b * b % md;
  }
  return ret;
}
const int MAX = 200110;
int pre[MAX], n, k;
pair<int, int> v[MAX];
vector<vector<int> > io(MAX);
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first >> v[i].second;
    pre[v[i].first]++;
    pre[v[i].second + 1]--;
    io[v[i].first].push_back(i + 1);
    io[v[i].second + 1].push_back(-i - 1);
  }
  for (int i = 1; i < MAX; ++i) {
    pre[i] += pre[i - 1];
  }
  vector<int> ans(MAX), sub(MAX);
  set<pair<int, int> > segments;
  int currentRemoved = 0;
  for (int i = 0; i < MAX; ++i) {
    currentRemoved += sub[i];
    for (int j = 0; j < io[i].size(); ++j) {
      if (io[i][j] > 0) {
        segments.insert(make_pair(v[io[i][j] - 1].second, io[i][j] - 1));
      } else {
        set<pair<int, int> >::iterator it =
            segments.find(make_pair(v[-io[i][j] - 1].second, -io[i][j] - 1));
        if (it != segments.end()) segments.erase(it);
      }
    }
    while (pre[i] - currentRemoved > k) {
      currentRemoved++;
      int maximalPositionReach = segments.rbegin()->second;
      segments.erase(prev(segments.end()));
      ans[maximalPositionReach] = 1;
      sub[v[maximalPositionReach].second + 1]--;
    }
  }
  int cnt = 0;
  for (int i = 0; i < ans.size(); ++i)
    if (ans[i]) cnt++;
  cout << cnt << '\n';
  for (int i = 0; i < ans.size(); ++i) {
    if (ans[i]) cout << i + 1 << ' ';
  }
  cout << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const long long md = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmd(long long a, long long b) {
  a = a % md;
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) {
      ret = ret * b % md;
    }
    b = b * b % md;
  }
  return ret;
}
const int MAX = 200110;
int pre[MAX], n, k;
pair<int, int> v[MAX];
vector<vector<int> > io(MAX);
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first >> v[i].second;
    pre[v[i].first]++;
    pre[v[i].second + 1]--;
    io[v[i].first].push_back(i + 1);
    io[v[i].second + 1].push_back(-i - 1);
  }
  for (int i = 1; i < MAX; ++i) {
    pre[i] += pre[i - 1];
  }
  vector<int> ans(MAX), sub(MAX);
  set<pair<int, int> > segments;
  int currentRemoved = 0;
  for (int i = 0; i < MAX; ++i) {
    currentRemoved += sub[i];
    for (int j = 0; j < io[i].size(); ++j) {
      if (io[i][j] > 0) {
        segments.insert(make_pair(v[io[i][j] - 1].second, io[i][j] - 1));
      } else {
        set<pair<int, int> >::iterator it =
            segments.find(make_pair(v[-io[i][j] - 1].second, -io[i][j] - 1));
        if (it != segments.end()) segments.erase(it);
      }
    }
    while (pre[i] - currentRemoved > k) {
      currentRemoved++;
      int maximalPositionReach = segments.rbegin()->second;
      segments.erase(prev(segments.end()));
      ans[maximalPositionReach] = 1;
      sub[v[maximalPositionReach].second + 1]--;
    }
  }
  int cnt = 0;
  for (int i = 0; i < ans.size(); ++i)
    if (ans[i]) cnt++;
  cout << cnt << '\n';
  for (int i = 0; i < ans.size(); ++i) {
    if (ans[i]) cout << i + 1 << ' ';
  }
  cout << '\n';
}

```