#include <bits/stdc++.h>
using namespace std;
pair<pair<int, int>, int> a[2000005];
int fw[4 * 2000005], n, k;
set<int> res;
set<pair<pair<int, int>, int> > l;
set<pair<pair<int, int>, int>, greater<pair<pair<int, int>, int> > > r;
void update(int x, int val) {
  for (; x < 2000005; x += x & (-x)) fw[x] += val;
}
int get(int x) {
  int res = 0;
  for (; x > 0; x -= x & (-x)) res += fw[x];
  return res;
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.second != b.first.second) return a.first.second < b.first.second;
  return a.first.first < b.first.first;
}
int main() {
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first.second >> a[i].first.first;
    a[i].second = i;
    update(a[i].first.first + 1, -1);
    update(a[i].first.second, 1);
  }
  sort(a + 1, a + 1 + n, cmp);
  int curr = 1;
  for (int i = 1; i < 2000005; i++) {
    while (a[curr].first.second == i) {
      r.insert(a[curr]);
      curr++;
    }
    while (get(i) > k) {
      auto temp = r.begin();
      update(temp->first.second, -1);
      update(temp->first.first + 1, 1);
      res.insert(temp->second);
      r.erase(temp);
    }
  }
  cout << res.size() << endl;
  for (auto i : res) cout << i << " ";
}

### Prompt

In Cpp, your task is to solve 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;
pair<pair<int, int>, int> a[2000005];
int fw[4 * 2000005], n, k;
set<int> res;
set<pair<pair<int, int>, int> > l;
set<pair<pair<int, int>, int>, greater<pair<pair<int, int>, int> > > r;
void update(int x, int val) {
  for (; x < 2000005; x += x & (-x)) fw[x] += val;
}
int get(int x) {
  int res = 0;
  for (; x > 0; x -= x & (-x)) res += fw[x];
  return res;
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.second != b.first.second) return a.first.second < b.first.second;
  return a.first.first < b.first.first;
}
int main() {
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first.second >> a[i].first.first;
    a[i].second = i;
    update(a[i].first.first + 1, -1);
    update(a[i].first.second, 1);
  }
  sort(a + 1, a + 1 + n, cmp);
  int curr = 1;
  for (int i = 1; i < 2000005; i++) {
    while (a[curr].first.second == i) {
      r.insert(a[curr]);
      curr++;
    }
    while (get(i) > k) {
      auto temp = r.begin();
      update(temp->first.second, -1);
      update(temp->first.first + 1, 1);
      res.insert(temp->second);
      r.erase(temp);
    }
  }
  cout << res.size() << endl;
  for (auto i : res) cout << i << " ";
}

```