#include <bits/stdc++.h>
using namespace std;
const double EPS = 0.0000000001;
const long long mod1 = 998244353;
const long long mod2 = 1000000007;
const long long mod3 = 1000000009;
const long long inf = 1000000000000000000;
using namespace std;
signed main() {
  long long n, a;
  cin >> n >> a;
  long long cnt[210];
  vector<pair<pair<long long, long long>, long long> > v;
  memset(cnt, 0, sizeof cnt);
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    v.push_back(make_pair(make_pair(l, r), i));
    for (long long z = l; z <= r; z++) cnt[z]++;
  }
  bool used[n];
  memset(used, 0, sizeof used);
  sort(v.begin(), v.end());
  vector<long long> res;
  for (long long z = 0; z < 210; z++) {
    while (cnt[z] > a) {
      long long ind = -1;
      long long sz = 0;
      for (long long j = 0; j < v.size(); j++) {
        if (v[j].first.second < z || v[j].first.first > z) continue;
        if (used[j]) continue;
        sz = max(sz, v[j].first.second);
        if (sz == (v[j].first.second)) {
          ind = j;
        }
      }
      used[ind] = 1;
      for (long long i = v[ind].first.first; i <= v[ind].first.second; i++)
        cnt[i]--;
      res.push_back(v[ind].second + 1);
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); i++) cout << res[i] << " ";
}

### 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 double EPS = 0.0000000001;
const long long mod1 = 998244353;
const long long mod2 = 1000000007;
const long long mod3 = 1000000009;
const long long inf = 1000000000000000000;
using namespace std;
signed main() {
  long long n, a;
  cin >> n >> a;
  long long cnt[210];
  vector<pair<pair<long long, long long>, long long> > v;
  memset(cnt, 0, sizeof cnt);
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    v.push_back(make_pair(make_pair(l, r), i));
    for (long long z = l; z <= r; z++) cnt[z]++;
  }
  bool used[n];
  memset(used, 0, sizeof used);
  sort(v.begin(), v.end());
  vector<long long> res;
  for (long long z = 0; z < 210; z++) {
    while (cnt[z] > a) {
      long long ind = -1;
      long long sz = 0;
      for (long long j = 0; j < v.size(); j++) {
        if (v[j].first.second < z || v[j].first.first > z) continue;
        if (used[j]) continue;
        sz = max(sz, v[j].first.second);
        if (sz == (v[j].first.second)) {
          ind = j;
        }
      }
      used[ind] = 1;
      for (long long i = v[ind].first.first; i <= v[ind].first.second; i++)
        cnt[i]--;
      res.push_back(v[ind].second + 1);
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); i++) cout << res[i] << " ";
}

```