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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
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...);
}
const int nax = 210;
int n, k;
vector<int> points(nax, 0);
bool check_valid() {
  for (int i = 1; i < nax; i++) {
    if (points[i] > k) return false;
  }
  return true;
}
int get_num_p(int l, int r) {
  int cnt = 0;
  for (int i = l; i <= r; i++) {
    if (points[i] > k) cnt++;
  }
  return cnt;
}
int main() {
  scanf("%d%d", &n, &k);
  vector<int> l(n);
  vector<int> r(n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i]; j++) points[j]++;
  };
  int cnt = 0;
  vector<int> inc(n, false);
  vector<int> ans;
  while (true) {
    bool okay = true;
    for (int i = 1; i < nax; i++) {
      if (points[i] <= k) continue;
      int max_seg = -1;
      int max_r = 0;
      for (int j = 0; j < n; j++) {
        if (inc[j]) continue;
        if (l[j] <= i && i <= r[j] && r[j] > max_r) {
          max_r = r[j];
          max_seg = j;
        }
      }
      assert(max_seg != -1);
      inc[max_seg] = true;
      for (int j = l[max_seg]; j <= r[max_seg]; j++) points[j]--;
      ;
      okay = false;
      ans.push_back(max_seg + 1);
      ;
      break;
    }
    if (okay) break;
  }
  cout << ans.size() << "\n";
  for (auto x : ans) cout << x << " ";
  cout << "\n";
  return 0;
}