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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = signed long long int;
const int N = 2e5 + 10;
struct seg {
  int l, r, i;
  inline bool operator<(const seg& o) const {
    if (l < o.l) {
      return true;
    }
    if (l > o.l) {
      return false;
    }
    if (r > o.r) {
      return true;
    }
    if (r < o.r) {
      return false;
    }
    return i > o.i;
  }
} a[N];
struct seg_by_ir {
  inline bool operator()(const seg& x, const seg& y) const {
    return x.r == y.r ? x < y : x.r > y.r;
  }
};
struct seg_by_r {
  inline bool operator()(const seg& x, const seg& y) const {
    return x.r == y.r ? x < y : x.r < y.r;
  }
};
set<seg, seg_by_ir> irq;
set<seg, seg_by_r> rq;
int n, k;
queue<seg> lq;
vector<seg> res;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    auto& cs = a[i];
    cin >> cs.l >> cs.r;
    cs.i = i + 1;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    lq.push(a[i]);
  }
  for (int i = 0; i < N; i++) {
    while (!rq.empty() && rq.begin()->r < i) {
      irq.erase(*rq.begin());
      rq.erase(rq.begin());
    }
    while (!lq.empty() && lq.front().l == i) {
      rq.insert(lq.front());
      irq.insert(lq.front());
      lq.pop();
    }
    while (irq.size() > k) {
      res.push_back(*irq.begin());
      rq.erase(*irq.begin());
      irq.erase(irq.begin());
    }
  }
  sort(res.begin(), res.end(), [](const seg& x, const seg& y) -> bool {
    return x.i == y.i ? x < y : x.i < y.i;
  });
  cout << res.size() << '\n';
  for (auto& x : res) {
    cout << x.i << ' ';
  }
}