#include <bits/stdc++.h>
using namespace std;
int p[200001]{}, k;
vector<int> R;
struct seg {
  int l, r, cb;
  seg(int ll, int rr) {
    l = ll;
    r = rr;
  }
  void calc() {
    cb = 0;
    for (int i = l; i <= r; ++i)
      if (p[i] > k) ++cb;
  }
  void rm() {
    for (int i = l; i <= r; ++i) --p[i];
    cb = l = r = 0;
  }
};
int main() {
  int n, l, r, m, i, j;
  cin >> n >> k;
  seg **ps = new seg *[n];
  for (i = 0; i < n; ++i) {
    cin >> l >> r;
    ps[i] = new seg(l, r);
    j = l;
    while (j <= r) ++p[j++];
  }
  for (i = 0; i < n; ++i) ps[i]->calc();
  for (i = 1; i < 200001; ++i)
    if (p[i] > k) {
      for (j = 0; j < n; ++j)
        if (i >= ps[j]->l && i <= ps[j]->r) {
          m = j;
          --ps[j]->cb;
          break;
        }
      for (++j; j < n; ++j)
        if (i >= ps[j]->l && i <= ps[j]->r) {
          --ps[j]->cb;
          if (ps[j]->cb > ps[m]->cb) m = j;
        }
      ps[m]->rm();
      R.emplace_back(m + 1);
      while (p[i] > k) {
        for (j = 0; j < n; ++j)
          if (i >= ps[j]->l && i <= ps[j]->r) {
            m = j;
            break;
          }
        for (++j; j < n; ++j)
          if (i >= ps[j]->l && i <= ps[j]->r) {
            if (ps[j]->cb > ps[m]->cb) m = j;
          }
        ps[m]->rm();
        R.emplace_back(m + 1);
      }
    }
  if (R.size() == 0) {
    cout << "0\n";
    return 0;
  }
  cout << R.size() << '\n' << R[0];
  for (i = 1; i < R.size(); ++i) cout << ' ' << R[i];
  cout << '\n';
  return 0;
}

### 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;
int p[200001]{}, k;
vector<int> R;
struct seg {
  int l, r, cb;
  seg(int ll, int rr) {
    l = ll;
    r = rr;
  }
  void calc() {
    cb = 0;
    for (int i = l; i <= r; ++i)
      if (p[i] > k) ++cb;
  }
  void rm() {
    for (int i = l; i <= r; ++i) --p[i];
    cb = l = r = 0;
  }
};
int main() {
  int n, l, r, m, i, j;
  cin >> n >> k;
  seg **ps = new seg *[n];
  for (i = 0; i < n; ++i) {
    cin >> l >> r;
    ps[i] = new seg(l, r);
    j = l;
    while (j <= r) ++p[j++];
  }
  for (i = 0; i < n; ++i) ps[i]->calc();
  for (i = 1; i < 200001; ++i)
    if (p[i] > k) {
      for (j = 0; j < n; ++j)
        if (i >= ps[j]->l && i <= ps[j]->r) {
          m = j;
          --ps[j]->cb;
          break;
        }
      for (++j; j < n; ++j)
        if (i >= ps[j]->l && i <= ps[j]->r) {
          --ps[j]->cb;
          if (ps[j]->cb > ps[m]->cb) m = j;
        }
      ps[m]->rm();
      R.emplace_back(m + 1);
      while (p[i] > k) {
        for (j = 0; j < n; ++j)
          if (i >= ps[j]->l && i <= ps[j]->r) {
            m = j;
            break;
          }
        for (++j; j < n; ++j)
          if (i >= ps[j]->l && i <= ps[j]->r) {
            if (ps[j]->cb > ps[m]->cb) m = j;
          }
        ps[m]->rm();
        R.emplace_back(m + 1);
      }
    }
  if (R.size() == 0) {
    cout << "0\n";
    return 0;
  }
  cout << R.size() << '\n' << R[0];
  for (i = 1; i < R.size(); ++i) cout << ' ' << R[i];
  cout << '\n';
  return 0;
}

```