#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int ox[N], sum[N];
struct node {
  int r, idx;
  node() {}
  node(int r, int idx) : r(r), idx(idx) {}
  bool operator<(const node a) const { return r < a.r; }
};
queue<node> q[N];
priority_queue<node> edge;
stack<int> s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int k, n, l, r, maxr = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    maxr = max(maxr, r);
    q[l].push(node(r, i));
    ++ox[l++];
    --ox[r + 1];
  }
  for (int i = 1; i <= maxr; i++) {
    sum[i] = sum[i - 1] + ox[i];
    while (!q[i].empty()) {
      edge.push(q[i].front());
      q[i].pop();
    }
    if (sum[i] <= k) continue;
    while (sum[i] > k) {
      s.push(edge.top().idx);
      sum[i]--;
      ++ox[edge.top().r + 1];
      edge.pop();
    }
  }
  cout << s.size() << endl;
  while (!s.empty()) {
    cout << s.top() << ' ';
    s.pop();
  }
}

### Prompt

Your task is to create 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 ≤ 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 + 10;
int ox[N], sum[N];
struct node {
  int r, idx;
  node() {}
  node(int r, int idx) : r(r), idx(idx) {}
  bool operator<(const node a) const { return r < a.r; }
};
queue<node> q[N];
priority_queue<node> edge;
stack<int> s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int k, n, l, r, maxr = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    maxr = max(maxr, r);
    q[l].push(node(r, i));
    ++ox[l++];
    --ox[r + 1];
  }
  for (int i = 1; i <= maxr; i++) {
    sum[i] = sum[i - 1] + ox[i];
    while (!q[i].empty()) {
      edge.push(q[i].front());
      q[i].pop();
    }
    if (sum[i] <= k) continue;
    while (sum[i] > k) {
      s.push(edge.top().idx);
      sum[i]--;
      ++ox[edge.top().r + 1];
      edge.pop();
    }
  }
  cout << s.size() << endl;
  while (!s.empty()) {
    cout << s.top() << ' ';
    s.pop();
  }
}

```