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;
using pii = pair<int, int>;
const int MAX = 200002;
int fenwik[MAX];
vector<pair<pii, int>> segs;
int used[MAX];
struct cmp {
  bool operator()(const pair<pii, int>& lhs, const pair<pii, int>& rhs) const {
    return lhs.first.second < rhs.first.second;
  }
};
bool cmp2(pair<pii, int> A, pair<pii, int> B) {
  return A.first.first != B.first.first ? A.first.first < B.first.first
                                        : A.first.second < B.first.second;
}
int sum(int n) {
  int ret = 0;
  while (n > 0) {
    ret += fenwik[n];
    n ^= (n & -n);
  }
  return ret;
}
void update(int n, int diff) {
  while (n <= MAX) {
    fenwik[n] += diff;
    n += (n & -n);
  }
}
int main() {
  cin.tie(0);
  ios_base ::sync_with_stdio(0);
  int N, M;
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    int a, b;
    cin >> a >> b;
    segs.push_back({{a, b}, i});
    update(a, 1);
    update(b + 1, -1);
  }
  vector<int> ans;
  sort(segs.begin(), segs.end(), cmp2);
  multiset<pair<pii, int>, cmp> cursegs;
  for (int i = 1, j = 0; i < 200001; i++) {
    int temp = sum(i);
    if (temp > M) {
      int d = temp - M;
      for (; j < segs.size(); j++) {
        if (segs[j].first.first <= i) {
          cursegs.insert(segs[j]);
        } else {
          break;
        }
      }
      while (d) {
        pair<pii, int> t = *(prev(cursegs.end()));
        cursegs.erase(prev(cursegs.end()));
        update(t.first.first, -1);
        update(t.first.second + 1, 1);
        ans.push_back(t.second + 1);
        d--;
      }
    }
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << ' ';
  }
}