#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k;
struct Node {
  int id, r;
  bool operator<(Node a) const { return r < a.r; }
};
Node a[500][500];
int cnt[500];
bool used[500];
int num[500];
int main() {
  cin >> n >> k;
  int x, y;
  for (int i = 1; i <= n; i++) {
    cin >> x >> y;
    for (int j = x; j <= y; j++) {
      cnt[j]++;
      a[j][cnt[j]].id = i;
      a[j][cnt[j]].r = y;
    }
  }
  for (int i = 1; i <= 200; i++) {
    sort(a[i] + 1, a[i] + 1 + cnt[i]);
    num[i] = cnt[i];
  }
  int ans = 0;
  for (int i = 1; i <= 200; i++) {
    if (num[i] > k) {
      while (num[i] > k) {
        if (used[a[i][cnt[i]].id]) {
          cnt[i]--;
          continue;
        }
        for (int j = i; j <= a[i][cnt[i]].r; j++) {
          num[j]--;
        }
        used[a[i][cnt[i]].id] = 1;
        cnt[i]--;
        ans++;
      }
    }
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    if (used[i]) {
      cout << i << " ";
    }
  }
}

### Prompt

Please formulate 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;
unsigned long long n, k;
struct Node {
  int id, r;
  bool operator<(Node a) const { return r < a.r; }
};
Node a[500][500];
int cnt[500];
bool used[500];
int num[500];
int main() {
  cin >> n >> k;
  int x, y;
  for (int i = 1; i <= n; i++) {
    cin >> x >> y;
    for (int j = x; j <= y; j++) {
      cnt[j]++;
      a[j][cnt[j]].id = i;
      a[j][cnt[j]].r = y;
    }
  }
  for (int i = 1; i <= 200; i++) {
    sort(a[i] + 1, a[i] + 1 + cnt[i]);
    num[i] = cnt[i];
  }
  int ans = 0;
  for (int i = 1; i <= 200; i++) {
    if (num[i] > k) {
      while (num[i] > k) {
        if (used[a[i][cnt[i]].id]) {
          cnt[i]--;
          continue;
        }
        for (int j = i; j <= a[i][cnt[i]].r; j++) {
          num[j]--;
        }
        used[a[i][cnt[i]].id] = 1;
        cnt[i]--;
        ans++;
      }
    }
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    if (used[i]) {
      cout << i << " ";
    }
  }
}

```