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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
struct seg {
  int l, r, id;
  bool operator<(const seg& a) const {
    if (l == a.l) {
      return r == a.r ? id < a.id : r < a.r;
    } else
      return l < a.l;
  }
} a[maxn];
int n, k;
int add[maxn];
priority_queue<seg> pq;
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].l >> a[i].r;
    a[i].id = i;
    ++add[a[i].l];
    --add[a[i].r + 1];
  }
  sort(a + 1, a + n + 1);
  int now = 0, ind = 1;
  for (int i = 1; i <= 200000; ++i) {
    while (ind <= n && a[ind].l == i) {
      int x = a[ind].l, y = a[ind].r;
      pq.push({y, x, a[ind].id});
      ind++;
    }
    now += add[i];
    while (now > k) {
      seg t = pq.top();
      pq.pop();
      if (t.l >= i) {
        ans.push_back(t.id);
        now--;
        add[t.l + 1]++;
      }
    }
    if (ind == n + 1) break;
  }
  cout << ans.size() << '\n';
  for (int i : ans) cout << i << ' ';
  cout << '\n';
  return 0;
}