#include <bits/stdc++.h>
using namespace std;
struct tpt {
  int val, id;
  tpt(int val = 0, int id = 0) : val(val), id(id) {}
  bool operator<(const tpt x) const {
    return val == x.val ? x.id < id : val > x.val;
  }
};
struct pt {
  int l, r, id;
} b[300005];
int n, m, a[300005], tans[300005], t;
set<tpt> st;
bool cmp(const pt x, const pt y) { return x.l == y.l ? x.r < y.r : x.l < y.l; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &b[i].l, &b[i].r), b[i].id = i;
  sort(b + 1, b + n + 1, cmp);
  int tnw = 1;
  for (int i = 1; i <= 200000; i++) {
    while (b[tnw].l == i) {
      while (st.size() && (--st.end())->val < i) {
        st.erase(--st.end());
      }
      st.insert(tpt(b[tnw].r, b[tnw].id));
      if (st.size() > m) {
        tans[++t] = st.begin()->id;
        st.erase(st.begin());
      }
      tnw++;
    }
  }
  printf("%d\n", t);
  for (int i = 1; i <= t; i++) printf("%d ", tans[i]);
  return 0;
}

### Prompt

Please formulate 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 tpt {
  int val, id;
  tpt(int val = 0, int id = 0) : val(val), id(id) {}
  bool operator<(const tpt x) const {
    return val == x.val ? x.id < id : val > x.val;
  }
};
struct pt {
  int l, r, id;
} b[300005];
int n, m, a[300005], tans[300005], t;
set<tpt> st;
bool cmp(const pt x, const pt y) { return x.l == y.l ? x.r < y.r : x.l < y.l; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &b[i].l, &b[i].r), b[i].id = i;
  sort(b + 1, b + n + 1, cmp);
  int tnw = 1;
  for (int i = 1; i <= 200000; i++) {
    while (b[tnw].l == i) {
      while (st.size() && (--st.end())->val < i) {
        st.erase(--st.end());
      }
      st.insert(tpt(b[tnw].r, b[tnw].id));
      if (st.size() > m) {
        tans[++t] = st.begin()->id;
        st.erase(st.begin());
      }
      tnw++;
    }
  }
  printf("%d\n", t);
  for (int i = 1; i <= t; i++) printf("%d ", tans[i]);
  return 0;
}

```