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 MAXN = 2e5 + 10;
int n, k, cnt[MAXN], Count, ans[MAXN], sum;
pair<int, int> p[MAXN];
vector<int> s[MAXN];
priority_queue<pair<int, int> > q;
int read() {
  int sss = 0, fff = 1;
  char ccc;
  ccc = getchar();
  while (ccc < '0' || ccc > '9') {
    if (ccc == '-') fff = -1;
    ccc = getchar();
  }
  while (ccc >= '0' && ccc <= '9')
    sss = (sss << 1) + (sss << 3) + (ccc ^ '0'), ccc = getchar();
  return sss * fff;
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) {
    p[i].first = read();
    p[i].second = read();
    cnt[p[i].first]++;
    cnt[p[i].second + 1]--;
    s[p[i].first].push_back(i);
  }
  for (int i = 1; i <= 200000; i++) {
    sum += cnt[i];
    for (int j = 0; j < s[i].size(); j++) {
      int x = s[i][j];
      q.push(make_pair(p[x].second + 1, x));
    }
    while (sum > k) {
      cnt[q.top().first]++;
      cnt[p[q.top().second].first]--;
      ans[q.top().second] = true;
      Count++;
      q.pop();
      sum--;
    }
  }
  printf("%d\n", Count);
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", i);
  return 0;
}