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 M = 1e9 + 7;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int f = 0;
  cerr << '{';
  for (auto& i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> B(n);
  vector<array<int, 3>> A;
  for (int i = 0; i < n; ++i) {
    cin >> B[i].first >> B[i].second;
    array<int, 3> temp = {B[i].first, 1, i};
    array<int, 3> temp2 = {B[i].second + 1, -1, i};
    A.push_back(temp);
    A.push_back(temp2);
  }
  sort(A.begin(), A.end());
  set<array<int, 2>> cur;
  n = A.size();
  int j = 0;
  int cnt = 0;
  vector<int> ans;
  for (int i = 1; i <= 2e5; ++i) {
    while (!cur.empty() && ((*cur.rbegin())[0] < i)) {
      cur.erase(*cur.rbegin());
    }
    while (j < n && (A[j][0] == i)) {
      if (A[j][1] == 1) {
        cur.insert({B[A[j][2]].second, A[j][2]});
      } else if (A[j][1] == -1) {
        cur.erase({B[A[j][2]].second, A[j][2]});
      }
      ++j;
    }
    while ((int)cur.size() > k) {
      ans.push_back((*cur.rbegin())[1]);
      cur.erase(*cur.rbegin());
      ++cnt;
    }
  }
  cout << cnt << "\n";
  for (int i : ans) {
    cout << i + 1 << " ";
  }
  cout << "\n";
  return 0;
}