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 int N = 2e5 + 10;
int n, k;
int m = 0;
int ans[N];
int tree[N * 4], tag[N * 4];
struct st {
  int l, r;
  int id;
} s[N];
bool cmp(st a, st b) {
  if (a.r != b.r) return a.r < b.r;
  return a.l < b.l;
}
void spread(int node, int l, int r) {
  if (tag[node]) {
    int leftNode = node * 2 + 1;
    int rightNode = node * 2 + 2;
    tag[leftNode] += tag[node];
    tag[rightNode] += tag[node];
    tree[leftNode] += tag[node];
    tree[rightNode] += tag[node];
    tag[node] = 0;
  }
}
void update(int node, int l, int r, int L, int R, int k) {
  if (L <= l && R >= r) {
    tag[node] += k;
    tree[node] += k;
    return;
  }
  spread(node, l, r);
  int mid = (l + r) / 2;
  int leftNode = node * 2 + 1;
  int rightNode = node * 2 + 2;
  if (mid >= L) update(leftNode, l, mid, L, R, k);
  if (mid < R) update(rightNode, mid + 1, r, L, R, k);
  tree[node] = max(tree[leftNode], tree[rightNode]);
}
int ask(int node, int l, int r, int L, int R) {
  if (L <= l && R >= r) return tree[node];
  spread(node, l, r);
  int mid = (l + r) / 2;
  int leftNode = node * 2 + 1;
  int rightNode = node * 2 + 2;
  int leftRes = 0, rightRes = 0;
  if (mid >= L) leftRes = ask(leftNode, l, mid, L, R);
  if (mid < R) rightRes = ask(rightNode, mid + 1, r, L, R);
  return max(leftRes, rightRes);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &s[i].l, &s[i].r);
    s[i].id = i + 1;
  }
  sort(s, s + n, cmp);
  for (int i = 0; i < n; i++) {
    int cnt = ask(0, 0, N, s[i].l, s[i].r);
    if (cnt < k)
      update(0, 0, N, s[i].l, s[i].r, 1);
    else
      ans[m++] = s[i].id;
  }
  cout << m << endl;
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}