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;
struct nd {
  int l, r, id;
  friend bool operator<(nd aa, nd bb) { return aa.r < bb.r; }
} a[200010];
priority_queue<nd> q;
int cmp(nd xx, nd yy) { return xx.l < yy.l; }
int n, k, c[200010], ans[200010];
int main() {
  scanf("%d%d", &n, &k);
  int mxr = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    c[a[i].l]++;
    c[a[i].r + 1]--;
    a[i].id = i;
    mxr = max(mxr, a[i].r);
  }
  int j = 1, cnt = 0;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= mxr; i++) {
    while (a[j].l <= i && j <= n) {
      q.push(a[j]);
      j++;
    }
    c[i] += c[i - 1];
    while (c[i] > k) {
      nd t = q.top();
      q.pop();
      c[i]--;
      c[t.r + 1]++;
      ans[++cnt] = t.id;
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  return 0;
}