#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x;
  int type;
  int ind;
  Point(const int x, const int type, const int ind)
      : x(x), type(type), ind(ind) {}
  bool operator<(Point another) const {
    if (x ^ another.x) return x < another.x;
    return type < another.type;
  }
};
const long long int MOD = 1e9 + 7;
const long long int MAXN = 1e6 + 1;
const long double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    ;
    vector<Point> points;
    vector<pair<int, int>> segs(n);
    for (int i = 0; i < (n); ++i) {
      int a, b;
      cin >> a >> b;
      ;
      points.push_back(Point(a, -1, i));
      points.push_back(Point(b, 1, i));
      segs[i].first = a;
      segs[i].second = b;
    }
    sort(points.begin(), points.end());
    vector<int> ans;
    multiset<pair<int, int>> open;
    for (auto p : points) {
      if (p.type == -1) {
        open.insert({segs[p.ind].second, p.ind});
        if (open.size() > k) {
          multiset<pair<int, int>>::iterator last = open.end();
          --last;
          ans.push_back(last->second + 1);
          open.erase({*last});
        }
      } else
        open.erase({segs[p.ind].second, p.ind});
    }
    cout << ans.size() << '\n';
    for (int i = 0; i < (ans.size()); ++i) cout << ans[i] << " ";
    ;
  }
  return 0;
}

### 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;
struct Point {
  int x;
  int type;
  int ind;
  Point(const int x, const int type, const int ind)
      : x(x), type(type), ind(ind) {}
  bool operator<(Point another) const {
    if (x ^ another.x) return x < another.x;
    return type < another.type;
  }
};
const long long int MOD = 1e9 + 7;
const long long int MAXN = 1e6 + 1;
const long double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    ;
    vector<Point> points;
    vector<pair<int, int>> segs(n);
    for (int i = 0; i < (n); ++i) {
      int a, b;
      cin >> a >> b;
      ;
      points.push_back(Point(a, -1, i));
      points.push_back(Point(b, 1, i));
      segs[i].first = a;
      segs[i].second = b;
    }
    sort(points.begin(), points.end());
    vector<int> ans;
    multiset<pair<int, int>> open;
    for (auto p : points) {
      if (p.type == -1) {
        open.insert({segs[p.ind].second, p.ind});
        if (open.size() > k) {
          multiset<pair<int, int>>::iterator last = open.end();
          --last;
          ans.push_back(last->second + 1);
          open.erase({*last});
        }
      } else
        open.erase({segs[p.ind].second, p.ind});
    }
    cout << ans.size() << '\n';
    for (int i = 0; i < (ans.size()); ++i) cout << ans[i] << " ";
    ;
  }
  return 0;
}

```