#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 7;
int N, K;
struct node {
  int r, id;
  node(int b = 0, int c = 0) : r(b), id(c) {}
  friend bool operator<(node e1, node e2) { return e1.r < e2.r; }
};
vector<node> add[maxN], del[maxN];
multiset<node> now;
multiset<node>::iterator it;
vector<int> vt;
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1, l, r; i <= N; i++) {
    scanf("%d%d", &l, &r);
    add[l].push_back(node(r, i));
    del[r + 1].push_back(node(r, i));
  }
  for (int i = 1; i <= 200000; i++) {
    int len = (int)add[i].size();
    for (int j = 0; j < len; j++) now.insert(add[i][j]);
    len = (int)del[i].size();
    for (int j = 0; j < len; j++) now.erase(del[i][j]);
    len = (int)now.size();
    if (len > K) {
      int det = len - K;
      while (det) {
        it = now.end();
        vt.push_back((--it)->id);
        now.erase(it);
        det--;
      }
    }
  }
  int len = (int)vt.size();
  printf("%d\n", len);
  for (int i = 0; i < len; i++) printf("%d ", vt[i]);
  printf("\n");
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int maxN = 2e5 + 7;
int N, K;
struct node {
  int r, id;
  node(int b = 0, int c = 0) : r(b), id(c) {}
  friend bool operator<(node e1, node e2) { return e1.r < e2.r; }
};
vector<node> add[maxN], del[maxN];
multiset<node> now;
multiset<node>::iterator it;
vector<int> vt;
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1, l, r; i <= N; i++) {
    scanf("%d%d", &l, &r);
    add[l].push_back(node(r, i));
    del[r + 1].push_back(node(r, i));
  }
  for (int i = 1; i <= 200000; i++) {
    int len = (int)add[i].size();
    for (int j = 0; j < len; j++) now.insert(add[i][j]);
    len = (int)del[i].size();
    for (int j = 0; j < len; j++) now.erase(del[i][j]);
    len = (int)now.size();
    if (len > K) {
      int det = len - K;
      while (det) {
        it = now.end();
        vt.push_back((--it)->id);
        now.erase(it);
        det--;
      }
    }
  }
  int len = (int)vt.size();
  printf("%d\n", len);
  for (int i = 0; i < len; i++) printf("%d ", vt[i]);
  printf("\n");
  return 0;
}

```