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 MAX = 3e5;
int seg[4 * MAX];
int lazy[4 * MAX];
void push(int k) {
  seg[2 * k] += lazy[k];
  seg[2 * k + 1] += lazy[k];
  lazy[2 * k] += lazy[k];
  lazy[2 * k + 1] += lazy[k];
  lazy[k] = 0;
}
void build(int a[], int v, int tl, int tr) {
  if (tl == tr) {
    seg[v] = a[tl];
    lazy[v] = a[tl];
  } else {
    int mid = (tl + tr) / 2;
    build(a, 2 * v, tl, mid);
    build(a, 2 * v + 1, mid + 1, tr);
    seg[v] = max(seg[2 * v], seg[2 * v + 1]);
  }
}
int getmax(int v, int tl, int tr, int l, int r) {
  if (l > r) return -1e9;
  if (tl == l && tr == r) return seg[v];
  push(v);
  int mid = (tl + tr) / 2;
  return max(getmax(2 * v, tl, mid, l, min(r, mid)),
             getmax(2 * v + 1, mid + 1, tr, max(l, mid + 1), r));
}
void update(int v, int tl, int tr, int l, int r, int val) {
  if (l > r) return;
  if (tl == l && tr == r) {
    seg[v] += val;
    lazy[v] += val;
  } else {
    int mid = (tl + tr) / 2;
    update(2 * v, tl, mid, l, min(r, mid), val);
    update(2 * v + 1, mid + 1, tr, max(l, mid + 1), r, val);
    seg[v] = max(seg[2 * v], seg[2 * v + 1]);
  }
}
int32_t main() {
  int n, k;
  cin >> n >> k;
  vector<pair<pair<int, int>, int>> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first.first >> v[i].first.second;
    v[i].second = i;
  }
  sort(v.begin(), v.end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         if (b.first.second != a.first.second)
           return a.first.second < b.first.second;
         else
           return a.first.first > b.first.first;
       });
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    int maxi = getmax(1, 0, 200010, v[i].first.first, v[i].first.second);
    if (maxi >= k) {
      ans.push_back(v[i].second);
      continue;
    } else {
      update(1, 0, 200010, v[i].first.first, v[i].first.second, 1);
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (auto x : ans) cout << x + 1 << " ";
  return 0;
}