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 hell = 1e9 + 7;
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
const long long N = 2e5 + 5;
long long fact[N], invfact[N];
void precompute() {
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] = fact[i] % hell;
  }
  invfact[N - 1] = power(fact[N - 1], hell - 2, hell);
  for (long long i = N - 2; i >= 0; i--) {
    invfact[i] = invfact[i + 1] * (i + 1);
    invfact[i] = invfact[i] % hell;
  }
}
long long nCr(long long x, long long y) {
  if (x < y) return 0;
  long long num = fact[x] % hell;
  num = num * invfact[y] % hell;
  num = num * invfact[x - y] % hell;
  return num;
}
bool removed[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<vector<long long>> v;
    vector<pair<long long, long long>> vec;
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      vec.push_back({x, y});
      v.push_back({x, 0, i});
      v.push_back({y, 1, i});
    }
    sort(v.begin(), v.end());
    set<pair<long long, long long>> curr;
    for (long long i = 0; i < 2 * n; i++) {
      long long dis = v[i][0];
      long long typ = v[i][1];
      long long ind = v[i][2];
      if (typ == 0) {
        curr.insert({vec[ind].second, ind});
        if (curr.size() > k) {
          auto p = prev(curr.end());
          ans.push_back(p->second);
          removed[p->second] = 1;
          curr.erase(p);
        }
      } else {
        if (removed[ind] == true) continue;
        if (curr.find({dis, ind}) != curr.end()) curr.erase({dis, ind});
      }
    }
    cout << ans.size() << "\n";
    for (auto x : ans) cout << x + 1 << " ";
    cout << "\n";
  }
  return 0;
}