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 = 200000 + 7;
int n, k;
struct str {
  int l, r, i;
} a[N];
multiset<str> tr;
int ans[N], p;
int operator<(const str &a, const str &b) {
  if (a.r == b.r) return a.l < b.l;
  return a.r < b.r;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i)
    scanf("%d%d", &a[i].l, &a[i].r), a[i].i = i, ++a[i].r;
  sort(a + 1, a + n + 1, [](str a, str b) { return a.l < b.l; });
  for (int i = 1, j; i <= n; i = j) {
    for (j = i; a[i].l == a[j].l; ++j) tr.insert(a[j]);
    while (!tr.empty() && (*tr.begin()).r <= a[i].l) tr.erase(tr.begin());
    while (tr.size() > k) {
      auto it = tr.end();
      --it;
      ans[++p] = (*it).i;
      tr.erase(it);
    }
  }
  printf("%d\n", p);
  for (int i = 1; i <= p; ++i) printf("%d ", ans[i]);
  return 0;
}