#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long double PI = 3.1415926535898;
const long long LLMAX = LLONG_MAX;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int num[203] = {0};
  int l[n + 1], r[n + 1];
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    num[l[i]] += 1;
    num[r[i] + 1] -= 1;
  }
  for (int i = 1; i <= 201; i++) num[i] += num[i - 1];
  set<int> s;
  for (int i = 1; i < 201; i++) {
    while (num[i] > k) {
      int index = -1, end = 0;
      for (int j = 0; j < n; j++) {
        auto it = s.find(j + 1);
        if (it == s.end()) {
          if (l[j] <= i && r[j] >= i && r[j] > end) {
            index = j;
            end = r[j];
          }
        }
      }
      for (int j = l[index]; j <= r[index]; j++) num[j]--;
      s.insert(index + 1);
    }
  }
  cout << s.size() << "\n";
  auto it = s.begin();
  for (it; it != s.end(); it++) cout << *it << " ";
}

### Prompt

Your challenge is to write 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 MOD = 1e9 + 7;
const long double PI = 3.1415926535898;
const long long LLMAX = LLONG_MAX;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int num[203] = {0};
  int l[n + 1], r[n + 1];
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    num[l[i]] += 1;
    num[r[i] + 1] -= 1;
  }
  for (int i = 1; i <= 201; i++) num[i] += num[i - 1];
  set<int> s;
  for (int i = 1; i < 201; i++) {
    while (num[i] > k) {
      int index = -1, end = 0;
      for (int j = 0; j < n; j++) {
        auto it = s.find(j + 1);
        if (it == s.end()) {
          if (l[j] <= i && r[j] >= i && r[j] > end) {
            index = j;
            end = r[j];
          }
        }
      }
      for (int j = l[index]; j <= r[index]; j++) num[j]--;
      s.insert(index + 1);
    }
  }
  cout << s.size() << "\n";
  auto it = s.begin();
  for (it; it != s.end(); it++) cout << *it << " ";
}

```