#include <bits/stdc++.h>
int n, k;
struct Segment {
  int l, r, id;
  bool operator<(const Segment& x) const {
    if (r == x.r) return id < x.id;
    return r > x.r;
  }
} se[200010];
std::vector<int> left;
inline bool cmpleft(const int& a, const int& b) { return se[a].l < se[b].l; }
std::vector<int> right;
inline bool cmpright(const int& a, const int& b) { return se[a].r < se[b].r; }
bool disp[200010];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, l, r; i <= n; i++) {
    scanf("%d%d", &l, &r);
    se[i].l = l;
    se[i].r = r;
    se[i].id = i;
    left.push_back(i);
    right.push_back(i);
  }
  std::sort(left.begin(), left.end(), cmpleft);
  std::sort(right.begin(), right.end(), cmpright);
  unsigned int pl = 0, pr = 0;
  std::set<Segment> curseg;
  int m = 0;
  for (int i = 1; i <= 200010; i++) {
    while (pl < left.size() && se[left[pl]].l <= i)
      curseg.insert(se[left[pl++]]);
    while (pr < right.size() & se[right[pr]].r < i)
      curseg.erase(*curseg.find(se[right[pr++]]));
    while (curseg.size() > k) {
      int id = curseg.begin()->id;
      curseg.erase(curseg.begin());
      disp[id] = 1;
      m++;
    }
  }
  printf("%d\n", m);
  for (int i = 1; i <= n; i++) {
    if (!disp[i]) continue;
    printf("%d ", i);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
int n, k;
struct Segment {
  int l, r, id;
  bool operator<(const Segment& x) const {
    if (r == x.r) return id < x.id;
    return r > x.r;
  }
} se[200010];
std::vector<int> left;
inline bool cmpleft(const int& a, const int& b) { return se[a].l < se[b].l; }
std::vector<int> right;
inline bool cmpright(const int& a, const int& b) { return se[a].r < se[b].r; }
bool disp[200010];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, l, r; i <= n; i++) {
    scanf("%d%d", &l, &r);
    se[i].l = l;
    se[i].r = r;
    se[i].id = i;
    left.push_back(i);
    right.push_back(i);
  }
  std::sort(left.begin(), left.end(), cmpleft);
  std::sort(right.begin(), right.end(), cmpright);
  unsigned int pl = 0, pr = 0;
  std::set<Segment> curseg;
  int m = 0;
  for (int i = 1; i <= 200010; i++) {
    while (pl < left.size() && se[left[pl]].l <= i)
      curseg.insert(se[left[pl++]]);
    while (pr < right.size() & se[right[pr]].r < i)
      curseg.erase(*curseg.find(se[right[pr++]]));
    while (curseg.size() > k) {
      int id = curseg.begin()->id;
      curseg.erase(curseg.begin());
      disp[id] = 1;
      m++;
    }
  }
  printf("%d\n", m);
  for (int i = 1; i <= n; i++) {
    if (!disp[i]) continue;
    printf("%d ", i);
  }
  return 0;
}

```