#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;
}
vector<int> ans;
vector<pair<int, int> > g[N];
int add[N << 2], mx[N << 2];
void pd(int rt) {
  if (add[rt]) {
    mx[rt << 1] += add[rt], add[rt << 1] += add[rt];
    mx[rt << 1 | 1] += add[rt], add[rt << 1 | 1] += add[rt];
    add[rt] = 0;
  }
}
void upd(int rt, int l, int r, int ql, int qr, int v) {
  if (ql <= l && r <= qr) {
    mx[rt] += v;
    add[rt] += v;
    return;
  }
  pd(rt);
  int mid = (l + r) >> 1;
  if (ql <= mid) upd(rt << 1, l, mid, ql, qr, v);
  if (qr > mid) upd(rt << 1 | 1, mid + 1, r, ql, qr, v);
  mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = int(1); i <= int(n); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    g[r].push_back(make_pair(l, i));
  }
  for (int i = int(1); i <= int(N - 1); ++i) {
    for (auto x : g[i]) {
      upd(1, 1, N, x.first, i, 1);
      if (mx[1] > k) upd(1, 1, N, x.first, i, -1), ans.push_back(x.second);
    }
  }
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
  return 0;
}

### Prompt

Generate a CPP solution to the following 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 

### Response

```cpp
#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;
}
vector<int> ans;
vector<pair<int, int> > g[N];
int add[N << 2], mx[N << 2];
void pd(int rt) {
  if (add[rt]) {
    mx[rt << 1] += add[rt], add[rt << 1] += add[rt];
    mx[rt << 1 | 1] += add[rt], add[rt << 1 | 1] += add[rt];
    add[rt] = 0;
  }
}
void upd(int rt, int l, int r, int ql, int qr, int v) {
  if (ql <= l && r <= qr) {
    mx[rt] += v;
    add[rt] += v;
    return;
  }
  pd(rt);
  int mid = (l + r) >> 1;
  if (ql <= mid) upd(rt << 1, l, mid, ql, qr, v);
  if (qr > mid) upd(rt << 1 | 1, mid + 1, r, ql, qr, v);
  mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = int(1); i <= int(n); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    g[r].push_back(make_pair(l, i));
  }
  for (int i = int(1); i <= int(N - 1); ++i) {
    for (auto x : g[i]) {
      upd(1, 1, N, x.first, i, 1);
      if (mx[1] > k) upd(1, 1, N, x.first, i, -1), ans.push_back(x.second);
    }
  }
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
  return 0;
}

```