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 N = 207;
int n, k, check[N], m[N], res;
vector<int> kq;
struct vcl {
  int first;
  int second;
  int vt;
};
bool cmp(vcl a, vcl b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
};
vcl a[N];
int bs(int l, int r, int x) {
  if (a[r].first <= x) return r;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (a[mid].first <= x && a[mid + 1].first > x)
      return mid;
    else if (a[mid].first <= x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
    m[a[i].first]++;
    m[a[i].second + 1]--;
    a[i].vt = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = 1; i <= a[n].second; i++) {
    m[i] += m[i - 1];
  }
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  for (int i = 1; i <= a[n].second; i++) {
    if (m[i] <= k) continue;
    int p = bs(1, n, i);
    int del = m[i] - k;
    for (int j = p; j >= 1; j--) {
      if (a[j].second >= i && check[j] == 0) {
        pq.push(pair<int, int>(a[j].second, j));
        if (pq.size() > del) pq.pop();
      }
    }
    while (pq.size() != 0) {
      int r = pq.top().first;
      int pos = pq.top().second;
      check[pos] = 1;
      for (int t = a[pos].first; t <= r; t++) {
        m[t]--;
      }
      pq.pop();
      res++;
      kq.push_back(a[pos].vt);
    }
  }
  cout << res << endl;
  for (int i = 0; i < kq.size(); i++) {
    cout << kq[i] << " ";
  }
}