#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7;
int n, k;
int l[250], r[250];
int cnt[250], CntRange[250];
vector<bool> vis;
int GetOptimal(int p) {
  pair<int, int> cur = make_pair(-1, -1);
  int pos = -1;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (p >= l[i] and p <= r[i] and (pos == -1 or cur.second < r[i]))
      cur = make_pair(l[i], r[i]), pos = i;
  }
  return pos;
}
int solve() {
  scanf("%d%d", &n, &k);
  vis.resize(n + 5, 0);
  int minleft = 1000, maxright = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    minleft = min(minleft, l[i]);
    maxright = max(maxright, r[i]);
  }
  for (int i = minleft; i <= maxright; i++)
    for (int j = 1; j <= n; j++) cnt[i] += (i >= l[j] and i <= r[j]);
  vector<int> Ans;
  for (int i = minleft; i <= maxright; i++) {
    while (cnt[i] > k) {
      int ret = GetOptimal(i);
      if (ret == -1) break;
      for (int j = l[ret]; j <= r[ret]; j++) cnt[j]--;
      vis[ret] = 1;
      Ans.push_back(ret);
    }
  }
  printf("%d\n", (int)Ans.size());
  for (int x : Ans) printf("%d ", x);
  return 0;
}
int main() { solve(); }

### Prompt

Develop a solution in cpp 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 int N = 2e5 + 5, mod = 1e9 + 7;
int n, k;
int l[250], r[250];
int cnt[250], CntRange[250];
vector<bool> vis;
int GetOptimal(int p) {
  pair<int, int> cur = make_pair(-1, -1);
  int pos = -1;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (p >= l[i] and p <= r[i] and (pos == -1 or cur.second < r[i]))
      cur = make_pair(l[i], r[i]), pos = i;
  }
  return pos;
}
int solve() {
  scanf("%d%d", &n, &k);
  vis.resize(n + 5, 0);
  int minleft = 1000, maxright = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    minleft = min(minleft, l[i]);
    maxright = max(maxright, r[i]);
  }
  for (int i = minleft; i <= maxright; i++)
    for (int j = 1; j <= n; j++) cnt[i] += (i >= l[j] and i <= r[j]);
  vector<int> Ans;
  for (int i = minleft; i <= maxright; i++) {
    while (cnt[i] > k) {
      int ret = GetOptimal(i);
      if (ret == -1) break;
      for (int j = l[ret]; j <= r[ret]; j++) cnt[j]--;
      vis[ret] = 1;
      Ans.push_back(ret);
    }
  }
  printf("%d\n", (int)Ans.size());
  for (int x : Ans) printf("%d ", x);
  return 0;
}
int main() { solve(); }

```