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>
const int maxn = 2e5 + 7;
const int inf = 0x3f3f3f3f;
using namespace std;
template <typename T>
inline void read(T &s) {
  s = 0;
  T w = 1, ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 1) + (s << 3) + (ch ^ 48);
    ch = getchar();
  }
  s *= w;
}
template <typename T>
inline void write(T s) {
  if (s < 0) putchar('-'), s = -s;
  if (s > 9) write(s / 10);
  putchar(s % 10 + '0');
}
int n;
int a[maxn];
struct tree {
  int l, r;
  long long dat, sum, add;
} t[maxn << 2];
struct str {
  int l, r, id;
} st[maxn];
void build(int p, int l, int r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    t[p].sum = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  t[p].dat = t[p << 1].dat + t[p << 1 | 1].dat;
}
void spread(int p) {
  if (t[p].add) {
    t[p << 1].add += t[p].add;
    t[p << 1 | 1].add += t[p].add;
    t[p << 1].dat += t[p].add;
    t[p << 1 | 1].dat += t[p].add;
    t[p].add = 0;
  }
}
void change(int p, int l, int r, int d) {
  if (l <= t[p].l && r >= t[p].r) {
    t[p].sum += (long long)d * (t[p].r - t[p].l + 1);
    t[p].add += d;
    t[p].dat += d;
    return;
  }
  spread(p);
  int mid = (t[p].l + t[p].r) >> 1;
  if (l <= mid) change(p << 1, l, r, d);
  if (r > mid) change(p << 1 | 1, l, r, d);
  t[p].sum = t[p << 1].sum + t[p << 1 | 1].sum;
  t[p].dat = max(t[p << 1].dat, t[p << 1 | 1].dat);
}
long long ask(int p, int l, int r) {
  if (l <= t[p].l && r >= t[p].r) return t[p].dat;
  spread(p);
  int mid = (t[p].l + t[p].r) >> 1;
  long long val = 0;
  if (l <= mid) val = max(val, ask(p << 1, l, r));
  if (r > mid) val = max(val, ask(p << 1 | 1, l, r));
  return val;
}
bool cmp(str a, str b) {
  if (a.r != b.r) {
    return a.r < b.r;
  } else {
    return a.l < b.l;
  }
}
int main() {
  int n, k;
  int t_maxn = 0;
  read(n);
  read(k);
  for (int i = 1; i <= n; i++) {
    read(st[i].l);
    read(st[i].r);
    t_maxn = max(t_maxn, st[i].r);
    st[i].id = i;
  }
  build(1, 1, t_maxn);
  sort(st + 1, st + 1 + n, cmp);
  vector<int> V;
  for (int i = 1; i <= n; i++) {
    int x, y;
    x = st[i].l;
    y = st[i].r;
    int temp_maxn = ask(1, x, y);
    if (temp_maxn >= k) {
      V.push_back(st[i].id);
    } else {
      change(1, x, y, 1);
    }
  }
  cout << V.size() << endl;
  for (int i = 0; i < V.size(); i++) {
    cout << V[i] << " ";
  }
  return 0;
}