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 long long M = 1e9 + 7;
long long IM = 1e16;
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << " ";
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void PTN(long long n = 1, long long a = 0, long long b = 0,
                long long c = 0) {
  if (n--) cout << a << " ";
  if (n--) cout << b << " ";
  if (n--) cout << c << " ";
  cout << "\n";
}
inline void op() {}
long long a[250] = {};
long long ck(long long k) {
  for (long long i = 1; i < 250; i++) {
    if (a[i] > k) return i;
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long i, j, k, x, y, u;
  {
    long long n, k;
    cin >> n >> k;
    vector<pair<long long, long long> > v;
    vector<long long> ans;
    for (i = 0; i < n; i++) {
      cin >> x >> y;
      a[x]++;
      a[y + 1]--;
      v.push_back({x, y});
    }
    for (i = 1; i < 250; i++) a[i] += a[i - 1];
    j = ck(k);
    while (j) {
      pair<long long, long long> in = {0, 0};
      long long l = 0;
      u = 0;
      for (i = 0; i < n; i++) {
        if (j >= v[i].first && j <= v[i].second) {
          if (v[i].second > l) {
            l = v[i].second;
            in = v[i];
            u = i;
          }
        }
      }
      for (i = in.first; i <= in.second; i++) a[i]--;
      ans.push_back(u + 1);
      v[u] = {0, 0};
      j = ck(k);
    }
    cout << (ans.size()) << "\n";
    PV(ans);
  }
  return 0;
}