#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9;
int seg[N * 4];
void update(int rt, int l, int r, int i, int j, int v) {
  if (l > j || r < i) return;
  if (l >= i && r <= j) {
    seg[rt] += v;
    return;
  }
  int lft = rt * 2;
  int rgt = lft + 1;
  int md = (l + r) / 2;
  seg[lft] += seg[rt];
  seg[rgt] += seg[rt];
  seg[rt] = 0;
  update(lft, l, md, i, j, v);
  update(rgt, md + 1, r, i, j, v);
}
int query(int rt, int l, int r, int id) {
  if (l > id || r < id) return 0;
  if (l == r) return seg[rt];
  int lft = rt * 2;
  int rgt = lft + 1;
  int md = (l + r) / 2;
  seg[lft] += seg[rt];
  seg[rgt] += seg[rt];
  seg[rt] = 0;
  return query(lft, l, md, id) + query(rgt, md + 1, r, id);
}
int main() {
  int n, k;
  cin >> n >> k;
  priority_queue<pair<int, pair<int, int> > > pq;
  vector<pair<int, pair<int, int> > > vc[N], sol;
  int p = 0;
  while (n--) {
    int l, r;
    cin >> l >> r;
    vc[l].push_back({l, {r, ++p}});
  }
  for (int i = 1; i < N; i++) {
    for (auto x : vc[i]) {
      update(1, 1, N - 5, x.first, x.second.first, +1);
      swap(x.first, x.second.first);
      pq.push(x);
    }
    while (query(1, 1, N - 5, i) > k) {
      auto x = pq.top();
      pq.pop();
      sol.push_back(x);
      update(1, 1, N - 5, x.second.first, x.first, -1);
    }
  }
  cout << sol.size() << "\n";
  for (auto x : sol) cout << x.second.second << " ";
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 ≤ 200) — 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 ≤ 200) — 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
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


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 int N = 2e5 + 9;
int seg[N * 4];
void update(int rt, int l, int r, int i, int j, int v) {
  if (l > j || r < i) return;
  if (l >= i && r <= j) {
    seg[rt] += v;
    return;
  }
  int lft = rt * 2;
  int rgt = lft + 1;
  int md = (l + r) / 2;
  seg[lft] += seg[rt];
  seg[rgt] += seg[rt];
  seg[rt] = 0;
  update(lft, l, md, i, j, v);
  update(rgt, md + 1, r, i, j, v);
}
int query(int rt, int l, int r, int id) {
  if (l > id || r < id) return 0;
  if (l == r) return seg[rt];
  int lft = rt * 2;
  int rgt = lft + 1;
  int md = (l + r) / 2;
  seg[lft] += seg[rt];
  seg[rgt] += seg[rt];
  seg[rt] = 0;
  return query(lft, l, md, id) + query(rgt, md + 1, r, id);
}
int main() {
  int n, k;
  cin >> n >> k;
  priority_queue<pair<int, pair<int, int> > > pq;
  vector<pair<int, pair<int, int> > > vc[N], sol;
  int p = 0;
  while (n--) {
    int l, r;
    cin >> l >> r;
    vc[l].push_back({l, {r, ++p}});
  }
  for (int i = 1; i < N; i++) {
    for (auto x : vc[i]) {
      update(1, 1, N - 5, x.first, x.second.first, +1);
      swap(x.first, x.second.first);
      pq.push(x);
    }
    while (query(1, 1, N - 5, i) > k) {
      auto x = pq.top();
      pq.pop();
      sol.push_back(x);
      update(1, 1, N - 5, x.second.first, x.first, -1);
    }
  }
  cout << sol.size() << "\n";
  for (auto x : sol) cout << x.second.second << " ";
}

```