#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
struct node {
  int l, r, id;
  friend bool operator<(node A, node B) {
    return A.r == B.r ? A.l < B.l : A.r < B.r;
  }
};
node a[N];
bool add[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll n, m;
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].l >> a[i].r;
    a[i].id = i;
  }
  sort(a + 1, a + 1 + n);
  multiset<int> b;
  for (int i = 0; i < m; ++i) b.insert(0);
  for (int i = 1; i <= n; ++i) {
    auto p = b.lower_bound(a[i].l);
    if (p != b.begin()) {
      add[a[i].id] = 1;
      --p;
      b.erase(p);
      b.insert(a[i].r);
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    if (!add[i]) ans.emplace_back(i);
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << ' ';
  cout << endl;
}

### Prompt

Your task is to create a Cpp solution 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;
using ll = long long;
const int N = 2e5 + 10;
struct node {
  int l, r, id;
  friend bool operator<(node A, node B) {
    return A.r == B.r ? A.l < B.l : A.r < B.r;
  }
};
node a[N];
bool add[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll n, m;
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].l >> a[i].r;
    a[i].id = i;
  }
  sort(a + 1, a + 1 + n);
  multiset<int> b;
  for (int i = 0; i < m; ++i) b.insert(0);
  for (int i = 1; i <= n; ++i) {
    auto p = b.lower_bound(a[i].l);
    if (p != b.begin()) {
      add[a[i].id] = 1;
      --p;
      b.erase(p);
      b.insert(a[i].r);
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    if (!add[i]) ans.emplace_back(i);
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << ' ';
  cout << endl;
}

```