#include <bits/stdc++.h>
const int maxn = 2e5 + 10;
const double eps = 1e-9;
struct Node {
  int y;
  int idex;
};
bool operator<(Node a, Node b) {
  if (a.y != b.y)
    return a.y < b.y;
  else
    return a.idex < b.idex;
}
std::vector<Node> g[maxn];
std::vector<int> out;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Node a;
    a.y = y;
    a.idex = i;
    g[x].push_back(a);
  }
  std::set<Node> st;
  for (int i = 1; i < maxn; i++) {
    while (st.size() && (*st.begin()).y < i) st.erase(*st.begin());
    for (int j = 0; j < g[i].size(); j++) {
      st.insert(g[i][j]);
    }
    while (st.size() > k) {
      out.push_back((*st.rbegin()).idex);
      st.erase(*st.rbegin());
    }
  }
  int len = out.size();
  printf("%d\n", len);
  for (int i = 0; i < len; i++) {
    printf("%d ", out[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
const int maxn = 2e5 + 10;
const double eps = 1e-9;
struct Node {
  int y;
  int idex;
};
bool operator<(Node a, Node b) {
  if (a.y != b.y)
    return a.y < b.y;
  else
    return a.idex < b.idex;
}
std::vector<Node> g[maxn];
std::vector<int> out;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Node a;
    a.y = y;
    a.idex = i;
    g[x].push_back(a);
  }
  std::set<Node> st;
  for (int i = 1; i < maxn; i++) {
    while (st.size() && (*st.begin()).y < i) st.erase(*st.begin());
    for (int j = 0; j < g[i].size(); j++) {
      st.insert(g[i][j]);
    }
    while (st.size() > k) {
      out.push_back((*st.rbegin()).idex);
      st.erase(*st.rbegin());
    }
  }
  int len = out.size();
  printf("%d\n", len);
  for (int i = 0; i < len; i++) {
    printf("%d ", out[i]);
  }
  printf("\n");
  return 0;
}

```