#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1);
const long long inf = 1LL << 30;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const int N = 200200;
set<pair<int, int>, greater<pair<int, int>>> s;
vector<pair<int, int>> vl[N], vr[N];
map<int, int> L, R;
vector<int> ans;
set<int> ren;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    vl[l].push_back({r, i});
    vr[r].push_back({r, i});
    ren.insert(l);
    ren.insert(r);
    L[l]++;
    R[r]++;
  }
  int cnt = 0;
  for (auto x : ren) {
    cnt += L[x];
    for (auto p : vl[x]) s.insert(p);
    while (cnt > k) {
      pair<int, int> p = *s.begin();
      s.erase(s.begin());
      ans.push_back(p.second);
      R[p.first]--;
      cnt--;
    }
    cnt -= R[x];
    for (auto p : vr[x])
      if (s.count(p)) s.erase(p);
  }
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
}

### Prompt

Your challenge is to write 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;
const long double pi = acos(-1);
const long long inf = 1LL << 30;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const int N = 200200;
set<pair<int, int>, greater<pair<int, int>>> s;
vector<pair<int, int>> vl[N], vr[N];
map<int, int> L, R;
vector<int> ans;
set<int> ren;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    vl[l].push_back({r, i});
    vr[r].push_back({r, i});
    ren.insert(l);
    ren.insert(r);
    L[l]++;
    R[r]++;
  }
  int cnt = 0;
  for (auto x : ren) {
    cnt += L[x];
    for (auto p : vl[x]) s.insert(p);
    while (cnt > k) {
      pair<int, int> p = *s.begin();
      s.erase(s.begin());
      ans.push_back(p.second);
      R[p.first]--;
      cnt--;
    }
    cnt -= R[x];
    for (auto p : vr[x])
      if (s.count(p)) s.erase(p);
  }
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
}

```