#include <bits/stdc++.h>
using namespace std;
const long long BIG = 1000000000000000000LL;
void solve();
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
const long long borne = 2 * (long long)(1e5) + 5;
long long gl[borne], gr[borne];
long long toRem[borne];
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > ev;
  for (long long i = (0); i < (n); ++i) {
    long long l, r;
    cin >> l >> r;
    long long x = i + 1;
    gl[x] = l;
    gr[x] = r;
    ev.push_back({l, x});
    ev.push_back({r + 1, -x});
    toRem[x] = 1;
  }
  sort(ev.begin(), ev.end());
  set<pair<long long, long long> > disp;
  long long cur = 0;
  vector<long long> choix;
  for (auto yolo : ev) {
    long long indSeg = yolo.second;
    if (indSeg > 0) {
      ++cur;
      disp.insert({gr[indSeg], indSeg});
      if (cur > k) {
        auto it = disp.end();
        --it;
        long long indRem = it->second;
        choix.push_back(indRem);
        toRem[indRem] = 0;
        --cur;
        disp.erase(it);
      }
    } else {
      indSeg = -indSeg;
      if (toRem[indSeg]) {
        --cur;
        disp.erase({gr[indSeg], indSeg});
      }
    }
  }
  sort(choix.begin(), choix.end());
  cout << choix.size() << "\n";
  for (long long x : choix) {
    cout << x << " ";
  }
  cout << "\n";
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long BIG = 1000000000000000000LL;
void solve();
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
const long long borne = 2 * (long long)(1e5) + 5;
long long gl[borne], gr[borne];
long long toRem[borne];
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > ev;
  for (long long i = (0); i < (n); ++i) {
    long long l, r;
    cin >> l >> r;
    long long x = i + 1;
    gl[x] = l;
    gr[x] = r;
    ev.push_back({l, x});
    ev.push_back({r + 1, -x});
    toRem[x] = 1;
  }
  sort(ev.begin(), ev.end());
  set<pair<long long, long long> > disp;
  long long cur = 0;
  vector<long long> choix;
  for (auto yolo : ev) {
    long long indSeg = yolo.second;
    if (indSeg > 0) {
      ++cur;
      disp.insert({gr[indSeg], indSeg});
      if (cur > k) {
        auto it = disp.end();
        --it;
        long long indRem = it->second;
        choix.push_back(indRem);
        toRem[indRem] = 0;
        --cur;
        disp.erase(it);
      }
    } else {
      indSeg = -indSeg;
      if (toRem[indSeg]) {
        --cur;
        disp.erase({gr[indSeg], indSeg});
      }
    }
  }
  sort(choix.begin(), choix.end());
  cout << choix.size() << "\n";
  for (long long x : choix) {
    cout << x << " ";
  }
  cout << "\n";
}

```