#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18 + 5LL;
const long long mod = 1e9 + 7LL;
void solve();
void ITO();
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ITO();
  long long t = 1;
  while (t--) solve();
  return 0;
}
const long long N = (2e5 + 10);
void solve() {
  long long n, k;
  cin >> n >> k;
  long long psum[N] = {};
  set<vector<long long>> s;
  set<vector<long long>> now;
  vector<long long> res;
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    psum[l]++;
    psum[r + 1]--;
    s.insert({l, r, i});
  }
  for (long long i = 1; i < N; i++) {
    psum[i] += psum[i - 1];
    while (!s.empty() and (*s.begin())[0] <= i) {
      now.insert({(*s.begin())[1], (*s.begin())[2]});
      s.erase(s.begin());
    }
    if (psum[i] <= k) continue;
    while (psum[i] > k) {
      long long r = (*now.rbegin())[0];
      res.push_back((*now.rbegin())[1] + 1);
      now.erase({r, res.back() - 1});
      psum[r + 1] += 1;
      psum[i]--;
    }
  }
  cout << res.size() << "\n";
  for (auto x : res) cout << x << " ";
  cout << "\n";
  return;
}
void ITO() {}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 long long inf = 1e18 + 5LL;
const long long mod = 1e9 + 7LL;
void solve();
void ITO();
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ITO();
  long long t = 1;
  while (t--) solve();
  return 0;
}
const long long N = (2e5 + 10);
void solve() {
  long long n, k;
  cin >> n >> k;
  long long psum[N] = {};
  set<vector<long long>> s;
  set<vector<long long>> now;
  vector<long long> res;
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    psum[l]++;
    psum[r + 1]--;
    s.insert({l, r, i});
  }
  for (long long i = 1; i < N; i++) {
    psum[i] += psum[i - 1];
    while (!s.empty() and (*s.begin())[0] <= i) {
      now.insert({(*s.begin())[1], (*s.begin())[2]});
      s.erase(s.begin());
    }
    if (psum[i] <= k) continue;
    while (psum[i] > k) {
      long long r = (*now.rbegin())[0];
      res.push_back((*now.rbegin())[1] + 1);
      now.erase({r, res.back() - 1});
      psum[r + 1] += 1;
      psum[i]--;
    }
  }
  cout << res.size() << "\n";
  for (auto x : res) cout << x << " ";
  cout << "\n";
  return;
}
void ITO() {}

```