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;
const long long INF = 1e18;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
long long qpow(long long x, long long y, long long MOD) {
  long long a = 1;
  while (y) {
    if (y & 1) a = a * x % MOD;
    x = x * x % MOD;
    y >>= 1;
  }
  return a;
}
struct seg {
  int l, r, id;
} a[N];
vector<int> ans;
int c[N], d[N], p[N];
bool cmp(seg A, seg B) { return A.l == B.l ? A.r < B.r : A.l < B.l; }
struct node {
  int x, id;
  bool operator<(const node& r) const { return x < r.x; }
};
int n, k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = int(1); i <= int(n); ++i) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].id = i;
    c[a[i].l]++, c[a[i].r + 1]--;
  }
  sort(a + 1, a + 1 + n, cmp);
  priority_queue<node> q;
  int res = 0, cur = 1;
  for (int i = int(1); i <= int(N - 1); ++i) {
    while (a[cur].l <= i && cur <= n) q.push(node{a[cur].r, a[cur].id}), cur++;
    d[i] = d[i - 1] + c[i];
    while (d[i] > k) {
      node tp = q.top();
      q.pop();
      ans.push_back(tp.id), res++;
      d[i]--;
      c[tp.x + 1]++;
    }
  }
  printf("%d\n", res);
  for (auto i : ans) printf("%d ", i);
  return 0;
}