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;
vector<int> u[210000], v[210000];
int k, n, num, l[210000], r[210000], ans;
bool b[210000], d[210000];
struct nod {
  int na;
  bool operator<(const nod &x) const { return r[na] < r[x.na]; }
} tn;
priority_queue<nod> q;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    u[l[i]].push_back(i);
    v[r[i] + 1].push_back(i);
  }
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < v[i].size(); j++)
      if (b[v[i][j]]) {
        b[v[i][j]] = 0;
        num--;
      }
    for (int j = 0; j < u[i].size(); j++) {
      b[u[i][j]] = 1;
      num++;
      q.push((nod){u[i][j]});
    }
    while (num > k) {
      tn = q.top();
      q.pop();
      if (!b[tn.na]) continue;
      b[tn.na] = 0;
      num--;
      ans++;
      d[tn.na] = 1;
    }
  }
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++)
    if (d[i]) printf("%d ", i);
  return 0;
}