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;
int main() {
  int n, i, j, k, a[200003][3], sum, b[200003], t = 0, f = 0, c[200003][3], p,
                                                s, maxf = 0;
  map<int, map<int, int>> m1;
  map<int, int>::reverse_iterator itr;
  map<int, map<int, int>>::iterator it;
  map<int, int>::iterator ritr;
  map<int, int> m3;
  map<int, map<int, int>> m2;
  map<int, map<int, int>> m4;
  vector<int> v;
  cin >> n;
  cin >> k;
  for (i = 0; i <= 200003; i++) {
    a[i][0] = 0, c[i][0] = 0;
    a[i][1] = 0, c[i][1] = 0;
    b[i] = 0;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    scanf("%d", &p);
    a[j - 1][0]++;
    a[p - 1][1]++;
    m1[j - 1][p - 1]++;
    m4[j - 1][i] = p - 1;
    if (p - 1 > maxf) maxf = p - 1;
  }
  for (i = 0; i <= maxf; i++) {
    t = t + a[i][0];
    b[i] = t;
    t = t - a[i][1];
  }
  t = 0;
  s = 0;
  for (i = 0; i <= maxf; i++) {
    if (m1.find(i) != m1.end()) {
      it = m1.find(i);
      for (ritr = it->second.begin(); ritr != it->second.end(); ritr++)
        m3[ritr->first] += ritr->second;
      it = m4.find(i);
      for (ritr = it->second.begin(); ritr != it->second.end(); ritr++)
        m2[ritr->second][ritr->first];
    }
    if (b[i] - t > k) {
      for (j = 0; j < b[i] - t - k; j++) {
        itr = m3.rbegin();
        s++;
        c[i][0]++;
        c[itr->first][1]++;
        p = m2[itr->first].begin()->first;
        m2[itr->first].erase(p);
        if (m2[itr->first].size() == 0) m2.erase(itr->first);
        v.push_back(p);
        itr->second--;
        if (itr->second == 0) m3.erase(itr->first);
      }
    }
    t = t + c[i][0] - c[i][1];
  }
  printf("%d\n", s);
  for (i = 0; i < v.size(); i++) printf("%d ", v[i] + 1);
  printf("\n");
}