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 maxn = 200000 + 10;
const long long mod = 1000000000 + 7;
struct node {
  int l, r, id;
} a[maxn];
bool cmp(node a, node b) {
  if (a.r == b.r) return a.l < b.l;
  return a.r < b.r;
}
int tr[maxn << 2], lazy[maxn << 2];
void build(int o, int l, int r) {
  if (l == r) return;
  int mid = l + r >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
}
void pushdown(int o) {
  if (lazy[o]) {
    lazy[o << 1] += lazy[o];
    lazy[o << 1 | 1] += lazy[o];
    tr[o << 1] += lazy[o];
    tr[o << 1 | 1] += lazy[o];
    lazy[o] = 0;
  }
}
void update(int o, int l, int r, int ql, int qr) {
  if (ql == l && qr == r) {
    tr[o]++, lazy[o]++;
    return;
  }
  pushdown(o);
  int mid = l + r >> 1;
  if (qr <= mid)
    update(o << 1, l, mid, ql, qr);
  else if (ql > mid)
    update(o << 1 | 1, mid + 1, r, ql, qr);
  else
    update(o << 1, l, mid, ql, mid),
        update(o << 1 | 1, mid + 1, r, mid + 1, qr);
  tr[o] = max(tr[o << 1], tr[o << 1 | 1]);
}
int query(int o, int l, int r, int ql, int qr) {
  if (ql == l && qr == r) return tr[o];
  pushdown(o);
  int mid = l + r >> 1;
  int ans = 0;
  if (qr <= mid)
    ans = query(o << 1, l, mid, ql, qr);
  else if (ql > mid)
    ans = query(o << 1 | 1, mid + 1, r, ql, qr);
  else
    ans = max(query(o << 1, l, mid, ql, mid),
              query(o << 1 | 1, mid + 1, r, mid + 1, qr));
  tr[o] = max(tr[o << 1], tr[o << 1 | 1]);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].l >> a[i].r;
    a[i].id = i;
    mx = max(mx, a[i].r);
  }
  vector<int> ans;
  sort(a + 1, a + n + 1, cmp);
  build(1, 1, mx);
  for (int i = 1; i <= n; i++) {
    if (query(1, 1, mx, a[i].l, a[i].r) < k) {
      update(1, 1, mx, a[i].l, a[i].r);
    } else {
      ans.push_back(a[i].id);
    }
  }
  cout << ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (auto x : ans) cout << x << " ";
  cout << endl;
  return 0;
}