#include <bits/stdc++.h>
using namespace std;
int n, k;
int l[2009], r[2009], a[2009], b[2009], c[2009];
int main() {
  int ans = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i]; j++) a[j]++;
  }
  for (int i = 1; i <= 200; i++) {
    if (a[i] > k) {
      ans++;
      b[ans] = i;
    }
  }
  int anss = 0;
  for (int i = 1; i <= ans; i++) {
    if (a[b[i]] <= k)
      continue;
    else
      while (a[b[i]] > k) {
        int maxx = 0, bj = 0;
        for (int j = 1; j <= n; j++)
          if (l[j] <= b[i] && r[j] > maxx) {
            maxx = r[j];
            bj = j;
          }
        l[bj] = 2000;
        r[bj] = 0;
        anss++;
        c[anss] = bj;
        for (int j = i; j <= ans; j++)
          if (b[j] <= maxx) a[b[j]]--;
      }
  }
  printf("%d\n", anss);
  for (int i = 1; i <= anss; i++) printf("%d ", c[i]);
  return 0;
}

### 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;
int n, k;
int l[2009], r[2009], a[2009], b[2009], c[2009];
int main() {
  int ans = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i]; j++) a[j]++;
  }
  for (int i = 1; i <= 200; i++) {
    if (a[i] > k) {
      ans++;
      b[ans] = i;
    }
  }
  int anss = 0;
  for (int i = 1; i <= ans; i++) {
    if (a[b[i]] <= k)
      continue;
    else
      while (a[b[i]] > k) {
        int maxx = 0, bj = 0;
        for (int j = 1; j <= n; j++)
          if (l[j] <= b[i] && r[j] > maxx) {
            maxx = r[j];
            bj = j;
          }
        l[bj] = 2000;
        r[bj] = 0;
        anss++;
        c[anss] = bj;
        for (int j = i; j <= ans; j++)
          if (b[j] <= maxx) a[b[j]]--;
      }
  }
  printf("%d\n", anss);
  for (int i = 1; i <= anss; i++) printf("%d ", c[i]);
  return 0;
}

```