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;
using ll = long long;
using ld = double;
const ll MODBASE = 1000000007LL;
const int MAXN = 200010;
const int MAXM = 1010;
const int MAXK = 110;
const int MAXQ = 200010;
struct Segment {
  int l, r, cs;
  Segment(int l, int r, int cs) : l(l), r(r), cs(cs) {}
};
bool operator<(Segment a, Segment b) {
  if (a.r == b.r) return a.l > b.l;
  return a.r < b.r;
}
vector<Segment> a, b[MAXN];
int n, f[MAXN], k;
vector<int> res;
bool choose[MAXN];
int Max[MAXN * 8], lazy[MAXN * 8];
void lazyUpd(int k, int l, int r) {
  if (lazy[k] == 0) return;
  Max[k] += lazy[k];
  if (l < r) {
    lazy[k * 2] += lazy[k];
    lazy[k * 2 + 1] += lazy[k];
  }
  lazy[k] = 0;
}
void update(int k, int l, int r, int u, int v, int gt) {
  lazyUpd(k, l, r);
  if (l > r || r < u || v < l) return;
  if (u <= l && r <= v) {
    lazy[k] = gt;
    lazyUpd(k, l, r);
    return;
  }
  int m = (l + r) >> 1;
  update(k * 2, l, m, u, v, gt);
  update(k * 2 + 1, m + 1, r, u, v, gt);
  Max[k] = max(Max[k * 2], Max[k * 2 + 1]);
}
int get(int k, int l, int r, int u, int v) {
  lazyUpd(k, l, r);
  if (l > r || r < u || v < l) return 0;
  if (u <= l && r <= v) return Max[k];
  int m = (l + r) >> 1;
  return max(get(k * 2, l, m, u, v), get(k * 2 + 1, m + 1, r, u, v));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = (1); i <= (n); i++) {
    int l, r;
    cin >> l >> r;
    a.emplace_back(Segment(l, r, i));
  }
  sort(a.begin(), a.end());
  for (int i = (0); i <= (n - 1); i++) b[a[i].r].emplace_back(a[i]);
  for (int i = (1); i <= (200000); i++) {
    for (int j = (0); j <= (int((b[i]).size()) - 1); j++) {
      Segment s = b[i][j];
      int e = get(1, 1, 200000, s.l, s.r);
      if (e >= k) break;
      choose[s.cs] = true;
      update(1, 1, 200000, s.l, s.r, 1);
    }
  }
  for (int i = (1); i <= (n); i++)
    if (!choose[i]) res.emplace_back(i);
  cout << int((res).size()) << "\n";
  for (int i = (0); i <= (int((res).size()) - 1); i++) cout << res[i] << ' ';
  return 0;
}