#include <bits/stdc++.h>
using namespace std;
signed main() {
  int n, k;
  cin >> n >> k;
  int seg[n + 1][2];
  int cnt[201];
  for (int i = 0; i <= 200; i++) cnt[i] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> seg[i][0] >> seg[i][1];
    for (int j = seg[i][0]; j <= seg[i][1]; j++) {
      cnt[j]++;
    }
  }
  vector<int> finall;
  while (1) {
    set<int> st;
    for (int i = 1; i <= 200; i++) {
      if (cnt[i] > k) st.insert(i);
    }
    if (st.size() == 0) break;
    int a = *(st.begin());
    int indexx = -1;
    int maxi = 0;
    for (int i = 1; i <= n; i++) {
      int aa = seg[i][0];
      int bb = seg[i][1];
      if (aa <= a) {
        int h = bb - a + 1;
        if (h > maxi) {
          maxi = h;
          indexx = i;
        }
      }
    }
    int l = seg[indexx][0];
    int r = seg[indexx][1];
    seg[indexx][0] = 200;
    seg[indexx][1] = -1;
    for (int i = l; i <= r; i++) cnt[i]--;
    finall.push_back(indexx);
  }
  sort(finall.begin(), finall.end());
  cout << finall.size() << endl;
  for (int i = 0; i < finall.size(); i++) cout << finall[i] << " ";
  return 0;
}

### Prompt

Please create a solution in cpp 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;
signed main() {
  int n, k;
  cin >> n >> k;
  int seg[n + 1][2];
  int cnt[201];
  for (int i = 0; i <= 200; i++) cnt[i] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> seg[i][0] >> seg[i][1];
    for (int j = seg[i][0]; j <= seg[i][1]; j++) {
      cnt[j]++;
    }
  }
  vector<int> finall;
  while (1) {
    set<int> st;
    for (int i = 1; i <= 200; i++) {
      if (cnt[i] > k) st.insert(i);
    }
    if (st.size() == 0) break;
    int a = *(st.begin());
    int indexx = -1;
    int maxi = 0;
    for (int i = 1; i <= n; i++) {
      int aa = seg[i][0];
      int bb = seg[i][1];
      if (aa <= a) {
        int h = bb - a + 1;
        if (h > maxi) {
          maxi = h;
          indexx = i;
        }
      }
    }
    int l = seg[indexx][0];
    int r = seg[indexx][1];
    seg[indexx][0] = 200;
    seg[indexx][1] = -1;
    for (int i = l; i <= r; i++) cnt[i]--;
    finall.push_back(indexx);
  }
  sort(finall.begin(), finall.end());
  cout << finall.size() << endl;
  for (int i = 0; i < finall.size(); i++) cout << finall[i] << " ";
  return 0;
}

```