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 PYTHON):

Solution:
from sys import stdin
import heapq
raw_input = lambda: stdin.readline().rstrip()
input = lambda: int(raw_input())
I=lambda: map(int, raw_input().split())
n,k = I()
lst = []
for i in xrange(n):
	l,r = I()
	lst.append((l, 0, i+1, r+1))
	lst.append((r+1, 1, i+1))
lst.sort(key=lambda x: x[0])
q = []
lLst = len(lst)
c = 0
j = 0
i = 0
rLst = []
s = set()
# print lst
while i<lLst:
	j = lst[i][0]
	w = j
	while i<lLst and j==lst[i][0]:
		if lst[i][1] == 1:
			w = lst[i][2]
			if w not in s:
				c -= 1
		else:
			_,_,w,r = lst[i]
			c += 1
			heapq.heappush(q, (-r, w))
		i += 1
	# print j, c, '-'
	while c>k:
		p = heapq.heappop(q)[1]
		s.add(p)
		rLst.append(p)
		c -= 1
print len(rLst)
print ' '.join([str(x) for x in rLst])