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 
#include <bits/stdc++.h>
using namespace std;
const long long N = 200010;
void solve() {
  long long n, m, i, j, k;
  cin >> n >> k;
  vector<pair<long long, long long> > a(n);
  vector<long long> f(1000), fr(1000);
  for (i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    f[a[i].first]++;
    f[a[i].second + 1]--;
  }
  for (i = 1; i <= 200; i++) fr[i] = fr[i - 1] + f[i];
  long long ans = 0;
  long long q = 10;
  vector<long long> v;
  vector<long long> vis(n + 1);
  while (1) {
    long long x = 0, mx = 0;
    for (i = 0; i < n; i++) {
      if (vis[i]) continue;
      long long l = a[i].first;
      long long r = a[i].second;
      long long c = 0;
      for (j = l; j <= r; j++) {
        if (fr[j] > k) {
          c++;
        }
      }
      if (c > mx) {
        mx = c;
        x = i;
      }
    }
    if (mx == 0) break;
    ans++;
    v.push_back(x + 1);
    vis[x] = 1;
    for (i = a[x].first; i <= a[x].second; i++) {
      fr[i]--;
    }
  }
  cout << ans << "\n";
  for (i = 0; i < v.size(); i++) cout << v[i] << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, i, j;
  long long q = 1;
  while (q--) solve();
}
