#include <bits/stdc++.h>
using namespace std;
int a[205] = {0}, c[205] = {0}, u = 1;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int b[205][2] = {0};
  int i, min = 1000000, max = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &b[i][0], &b[i][1]);
    if (min > b[i][0]) min = b[i][0];
    if (max < b[i][1]) max = b[i][1];
    for (int j = b[i][0]; j <= b[i][1]; j++) a[j]++;
  }
  int ans = 0;
  for (i = min; i <= max; i++) {
    if (a[i] > k)
      while (a[i] != k) {
        int delete_number, maxlength = 0;
        for (int j = 1; j <= n; j++) {
          if (i >= b[j][0] && i <= b[j][1]) {
            if (maxlength <= b[j][1] - i) {
              maxlength = b[j][1] - i;
              delete_number = j;
            }
          }
        }
        for (int j = i; j - i <= maxlength; j++) {
          a[j]--;
        }
        ans++;
        c[u] = delete_number;
        u++;
        b[delete_number][0] = b[delete_number][1] = -1;
      }
  }
  if (ans) {
    printf("%d\n", ans);
    for (i = 1; i < u; i++) printf("%d ", c[i]);
  } else
    printf("0\n");
  return 0;
}

### 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;
int a[205] = {0}, c[205] = {0}, u = 1;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int b[205][2] = {0};
  int i, min = 1000000, max = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &b[i][0], &b[i][1]);
    if (min > b[i][0]) min = b[i][0];
    if (max < b[i][1]) max = b[i][1];
    for (int j = b[i][0]; j <= b[i][1]; j++) a[j]++;
  }
  int ans = 0;
  for (i = min; i <= max; i++) {
    if (a[i] > k)
      while (a[i] != k) {
        int delete_number, maxlength = 0;
        for (int j = 1; j <= n; j++) {
          if (i >= b[j][0] && i <= b[j][1]) {
            if (maxlength <= b[j][1] - i) {
              maxlength = b[j][1] - i;
              delete_number = j;
            }
          }
        }
        for (int j = i; j - i <= maxlength; j++) {
          a[j]--;
        }
        ans++;
        c[u] = delete_number;
        u++;
        b[delete_number][0] = b[delete_number][1] = -1;
      }
  }
  if (ans) {
    printf("%d\n", ans);
    for (i = 1; i < u; i++) printf("%d ", c[i]);
  } else
    printf("0\n");
  return 0;
}

```