#include <bits/stdc++.h>
using namespace std;
long long n, k, f[200009], z, ans, v[200009];
struct st {
  int l, r, i;
} a[200009];
vector<pair<int, int>> l[200009];
set<pair<int, int>> s;
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &a[i].l, &a[i].r);
    a[i].i = i;
    f[a[i].l]++, f[a[i].r + 1]--;
    l[a[i].l].push_back({a[i].r, a[i].i});
  }
  for (int i = 1; i <= 200002; i++) {
    f[i] += f[i - 1];
    for (int j = 0; j < l[i].size(); j++)
      s.insert({l[i][j].first, l[i][j].second});
    if (f[i] <= k) continue;
    z = f[i] - k;
    while (z--) {
      v[ans] = (*--s.end()).second;
      ans++;
      f[i + 1]--;
      f[(*--s.end()).first + 1]++;
      s.erase(--s.end());
    }
  }
  printf("%I64d\n", ans);
  for (int i = 0; i < ans; i++) {
    printf("%I64d ", v[i]);
  }
}

### 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;
long long n, k, f[200009], z, ans, v[200009];
struct st {
  int l, r, i;
} a[200009];
vector<pair<int, int>> l[200009];
set<pair<int, int>> s;
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &a[i].l, &a[i].r);
    a[i].i = i;
    f[a[i].l]++, f[a[i].r + 1]--;
    l[a[i].l].push_back({a[i].r, a[i].i});
  }
  for (int i = 1; i <= 200002; i++) {
    f[i] += f[i - 1];
    for (int j = 0; j < l[i].size(); j++)
      s.insert({l[i][j].first, l[i][j].second});
    if (f[i] <= k) continue;
    z = f[i] - k;
    while (z--) {
      v[ans] = (*--s.end()).second;
      ans++;
      f[i + 1]--;
      f[(*--s.end()).first + 1]++;
      s.erase(--s.end());
    }
  }
  printf("%I64d\n", ans);
  for (int i = 0; i < ans; i++) {
    printf("%I64d ", v[i]);
  }
}

```