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 int mod = 998244353;
long double eps = 1e-12;
long long int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
long long int n, res = 0, k = 0, h, ans = 0, m;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int x, long long int y) {
  return (x * y) / gcd(x, y);
}
long long int pot(long long int a, long long int b) {
  long long int r = 1;
  while (b) {
    if (b % 2) r = (r * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return r % mod;
}
void pr(long long int a[], long long int n) {
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << '\n';
}
void read(long long int a[], long long int n) {
  for (long long int i = 1; i <= n; i++) cin >> a[i];
}
void yes() { cout << "Yes\n"; }
void no() { cout << "No\n"; }
vector<long long int> v[300015];
void add_edge(long long int x, long long int y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
string s2, ch = "", s, s1 = "", t;
vector<pair<string, long long int> > vs;
vector<pair<pair<long long int, long long int>, long long int> > vvv, vvv2;
vector<pair<long long int, long long int> > vv[300015], vv2;
long long int a[300015] = {0}, b[300015] = {0};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int i = 0, x = 0, z = 0, y = 0, j = 0, q, mx = 0, mm, idx = 0,
                ok = 0, l = 0, r, negatif = 0, positif = 0;
  set<pair<long long int, long long int> >::iterator itw, ith;
  multiset<pair<long long int, long long int> >::iterator it;
  multiset<pair<long long int, long long int> > mst, mst2;
  cin >> n >> k;
  for (i = 1; i <= (long long int)(n); i++) {
    cin >> x >> y;
    a[x]++;
    a[y + 1]--;
    vvv.push_back({{x, y}, i});
  }
  sort(vvv.begin(), vvv.end());
  a[0] = 0;
  for (i = 1; i <= (long long int)(2e5); i++) {
    a[i] += a[i - 1];
  }
  long long int sm = 0;
  j = 0;
  vector<long long int> cur;
  for (i = 1; i <= 2e5; i++) {
    sm += b[i];
    while (j < vvv.size()) {
      l = vvv[j].first.first;
      r = vvv[j].first.second;
      idx = vvv[j].second;
      if (l > i) break;
      j++;
      mst.insert({-r, idx});
    }
    while ((sm + a[i]) > k) {
      it = mst.begin();
      r = abs((*it).first);
      idx = (*it).second;
      cur.push_back(idx);
      sm--;
      b[r + 1]++;
      mst.erase(it);
    }
  }
  cout << cur.size() << endl;
  if (cur.size())
    for (i = 0; i < cur.size(); i++) cout << cur[i] << " ";
  return 0;
}