#include <bits/stdc++.h>
using namespace std;
struct node {
  int s;
  int e;
  int id;
} a[200005];
int cmp(node p, node q) { return p.s < q.s; }
vector<int> ans;
set<pair<int, int> > st;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int maxn, minn;
  pair<int, int> tmp;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].s, &a[i].e);
    maxn = max(maxn, a[i].e);
    minn = min(minn, a[i].s);
    a[i].id = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  int pos = 1;
  for (int i = minn; i <= maxn; i++) {
    while (pos <= n && a[pos].s <= i) {
      st.insert(make_pair(a[pos].e, a[pos].id));
      pos++;
    }
    while (st.size() && st.begin()->first < i) {
      st.erase(st.begin());
    }
    while (st.size() > k) {
      tmp = *(--st.end());
      ans.push_back(tmp.second);
      st.erase(tmp);
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d ", v);
  }
  printf("\n");
}

### Prompt

Your task is to create 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;
struct node {
  int s;
  int e;
  int id;
} a[200005];
int cmp(node p, node q) { return p.s < q.s; }
vector<int> ans;
set<pair<int, int> > st;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int maxn, minn;
  pair<int, int> tmp;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].s, &a[i].e);
    maxn = max(maxn, a[i].e);
    minn = min(minn, a[i].s);
    a[i].id = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  int pos = 1;
  for (int i = minn; i <= maxn; i++) {
    while (pos <= n && a[pos].s <= i) {
      st.insert(make_pair(a[pos].e, a[pos].id));
      pos++;
    }
    while (st.size() && st.begin()->first < i) {
      st.erase(st.begin());
    }
    while (st.size() > k) {
      tmp = *(--st.end());
      ans.push_back(tmp.second);
      st.erase(tmp);
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d ", v);
  }
  printf("\n");
}

```