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 

Solution (in JAVA):

Solution:
import java.util.*;

public class D {

	public static void main(String[] args) {
		Map<Integer, List<int[]>> left2Seg = new HashMap<>(200010);
		Map<Integer, List<int[]>> right2Seg = new HashMap<>(200010);
		Set<Integer> curSegIds = new HashSet<>(200010);
		List<int[]> curSegs = new ArrayList<>();
		StringBuilder sb = new StringBuilder();

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), k = sc.nextInt(), max = 0;
		List<int[]> segments = new ArrayList<>(n);
		for (int i = 0, l, r; i < n; ++i) {
			l = sc.nextInt();
			r = sc.nextInt();
			int[] seg = {i + 1, l, r};
			segments.add(seg);
			max = Math.max(max, r);
		}
		sc.close();

		segments.sort((s1, s2) -> s1[1] != s2[1] ? s1[1] - s2[1] : s2[2] - s1[2]);
		segments.forEach(seg -> {
			left2Seg.putIfAbsent(seg[1], new ArrayList<>());
			left2Seg.get(seg[1]).add(seg);
			right2Seg.putIfAbsent(seg[2] + 1, new ArrayList<>());
			right2Seg.get(seg[2] + 1).add(seg);
		});

		int cnt = 0, removeCnt = 0;
		for (int i = 0; i <= max; ++i) {
			List<int[]> opens = left2Seg.get(i), closes = right2Seg.get(i);
			if (closes != null) {
				for (int[] seg : closes) {
					if (curSegIds.contains(seg[0])) {
						curSegIds.remove(seg[0]);
						--cnt;
					}
				}
			}
			if (opens != null) {
				for (int[] seg : opens) {
					curSegIds.add(seg[0]);
					curSegs.add(seg);
					++cnt;
				}
				if (cnt > k) {
					curSegs.sort((s1, s2) -> s1[2] - s2[2]);
					for (int j = curSegs.size() - 1; cnt > k && j >= 0; --j) {
						int[] removeSeg = curSegs.get(j);
						if (curSegIds.contains(removeSeg[0])) {
							curSegIds.remove(removeSeg[0]);
							--cnt;
							sb.append(removeSeg[0]).append(' ');
							++removeCnt;
						}
						curSegs.remove(j);
					}
				}
			}
		}

		System.out.println(removeCnt);
		System.out.println(sb.toString());
	}

}