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;
long long st[800010], lazy[800010], n;
set<pair<pair<int, int>, int> > A, B;
void update(int s, int e, int value, int l = 1, int r = 200000, int i = 0) {
  if (lazy[i] != 0) {
    long long x = (r - l + 1) * lazy[i];
    st[i] += x;
    if (l != r) {
      lazy[i * 2 + 1] += lazy[i];
      lazy[i * 2 + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  if (r < s || l > e) return;
  if (s <= l && r <= e) {
    long long x = (r - l + 1) * value;
    st[i] += x;
    if (l != r) {
      lazy[i * 2 + 1] += value;
      lazy[i * 2 + 2] += value;
    }
    return;
  }
  int mid = (l + r) >> 1;
  update(s, e, value, l, mid, i * 2 + 1);
  update(s, e, value, mid + 1, r, i * 2 + 2);
  st[i] = st[i * 2 + 1] + st[i * 2 + 2];
}
long long sum(int s, int e, int l = 1, int r = 200000, int i = 0) {
  if (lazy[i] != 0) {
    long long x = (r - l + 1) * lazy[i];
    st[i] += x;
    if (l != r) {
      lazy[i * 2 + 1] += lazy[i];
      lazy[i * 2 + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  if (r < s || l > e) return 0;
  if (s <= l && r <= e) return st[i];
  int mid = (l + r) >> 1;
  long long L = sum(s, e, l, mid, i * 2 + 1);
  long long R = sum(s, e, mid + 1, r, i * 2 + 2);
  return L + R;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int segments, k;
  cin >> segments >> k;
  vector<vector<pair<int, int> > > V(200000);
  for (int i = 0; i < segments; i++) {
    int l, r;
    cin >> l >> r;
    V[l].push_back({r, i + 1});
  }
  vector<int> ans;
  for (int i = 1; i <= 200000; i++) {
    while (A.size()) {
      pair<pair<int, int>, int> x = *A.begin();
      if (x.first.first < i) {
        A.erase({{x.first.first, x.first.second}, x.second});
        B.erase({{-x.first.first, x.first.second}, x.second});
      } else {
        break;
      }
    }
    for (int j = 0; j < V[i].size(); j++) {
      update(i, V[i][j].first, 1);
      A.insert({{V[i][j].first, i}, V[i][j].second});
      B.insert({{-V[i][j].first, i}, V[i][j].second});
    }
    long long diff = 0;
    diff = max(0LL, sum(i, i) - k);
    while (diff > 0) {
      diff--;
      pair<pair<int, int>, int> x = *B.begin();
      update(x.first.second, -x.first.first, -1);
      ans.push_back(x.second);
      A.erase({{-x.first.first, x.first.second}, x.second});
      B.erase({{x.first.first, x.first.second}, x.second});
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << "\n";
  return 0;
}