#include <bits/stdc++.h>
using namespace std;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1LL) {
    if (k & 1LL) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<int> st[3 * 100001];
set<pair<int, int> > curr;
vector<int> ind[3 * 100001];
void solve() {
  int n, k;
  cin >> n >> k;
  int A[3 * 100001 + 2];
  memset(A, 0, sizeof(A));
  int a, b;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    st[a].push_back(b);
    ind[a].push_back(i + 1);
  }
  vector<int> ans;
  for (int i = 1; i < 3 * 100001; i++) {
    A[i] += A[i - 1];
    for (int j = 0; j < st[i].size(); j++) {
      curr.insert(make_pair(st[i][j], ind[i][j]));
      A[i]++;
      A[st[i][j] + 1]--;
    }
    while (A[i] > k) {
      auto it = curr.end();
      it--;
      A[i]--;
      A[(it->first) + 1]++;
      ans.push_back(it->second);
      curr.erase(it);
    }
  }
  cout << ans.size() << '\n';
  sort((ans).begin(), (ans).end());
  for (int i : ans) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Generate a cpp solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1LL) {
    if (k & 1LL) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<int> st[3 * 100001];
set<pair<int, int> > curr;
vector<int> ind[3 * 100001];
void solve() {
  int n, k;
  cin >> n >> k;
  int A[3 * 100001 + 2];
  memset(A, 0, sizeof(A));
  int a, b;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    st[a].push_back(b);
    ind[a].push_back(i + 1);
  }
  vector<int> ans;
  for (int i = 1; i < 3 * 100001; i++) {
    A[i] += A[i - 1];
    for (int j = 0; j < st[i].size(); j++) {
      curr.insert(make_pair(st[i][j], ind[i][j]));
      A[i]++;
      A[st[i][j] + 1]--;
    }
    while (A[i] > k) {
      auto it = curr.end();
      it--;
      A[i]--;
      A[(it->first) + 1]++;
      ans.push_back(it->second);
      curr.erase(it);
    }
  }
  cout << ans.size() << '\n';
  sort((ans).begin(), (ans).end());
  for (int i : ans) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```