#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
const long double pi = acos(-1.0);
const long double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
multiset<tuple<int, int, int>> s;
tuple<int, int, int> a[200010];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = (1); i <= (n); i++) {
    int l, r;
    cin >> l >> r;
    a[i] = tie(l, r, i);
  }
  sort(a + 1, a + 1 + n);
  int p = 1;
  vector<int> ans;
  for (int i = (1); i <= (200000); i++) {
    while (!s.empty() && get<0>(*s.begin()) < i) s.erase(s.begin());
    while (p <= n && get<0>(a[p]) <= i) {
      s.emplace(get<1>(a[p]), get<0>(a[p]), get<2>(a[p]));
      p++;
    }
    while (s.size() > k) {
      auto it = s.end();
      it--;
      ans.emplace_back(get<2>(*it));
      s.erase(it);
    }
  }
  cout << ans.size() << '\n';
  for (auto it : ans) cout << it << " ";
}

### Prompt

In cpp, your task is to solve 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>
#pragma warning(disable : 4996)
using namespace std;
const long double pi = acos(-1.0);
const long double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
multiset<tuple<int, int, int>> s;
tuple<int, int, int> a[200010];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = (1); i <= (n); i++) {
    int l, r;
    cin >> l >> r;
    a[i] = tie(l, r, i);
  }
  sort(a + 1, a + 1 + n);
  int p = 1;
  vector<int> ans;
  for (int i = (1); i <= (200000); i++) {
    while (!s.empty() && get<0>(*s.begin()) < i) s.erase(s.begin());
    while (p <= n && get<0>(a[p]) <= i) {
      s.emplace(get<1>(a[p]), get<0>(a[p]), get<2>(a[p]));
      p++;
    }
    while (s.size() > k) {
      auto it = s.end();
      it--;
      ans.emplace_back(get<2>(*it));
      s.erase(it);
    }
  }
  cout << ans.size() << '\n';
  for (auto it : ans) cout << it << " ";
}

```