D1. Too Many Segments (easy version)
greedy
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 \([l_i; r_i]\) (\(l_i \le r_i\)) and it covers all integer points \(j\) such that \(l_i \le j \le r_i\).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.
The first line of the input contains two integers \(n\) and \(k\) (\(1 \le k \le n \le 200\)) — 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 \(r_i\) (\(1 \le l_i \le r_i \le 200\)) — the endpoints of the \(i\)-th segment.
In the first line print one integer \(m\) (\(0 \le m \le n\)) — the minimum number of segments you need to remove so that there are no bad points.In the second line print \(m\) distinct integers \(p_1, p_2, \dots, p_m\) (\(1 \le p_i \le n\)) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.