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 PYTHON3):

Solution:
from sys import stdin
import heapq

n,k = [int(x) for x in stdin.readline().split()]

segs = []

axis = [0 for x in range(200)]

for seg in range(n):
  l,r = [int(x) for x in stdin.readline().split()]
  l -= 1

  segs.append((l,r,seg+1))

segs.sort()

q = []

dummy = []

for l,r,ind in segs:
  heapq.heappush(q,(-r,l,ind))
  chill = True
  for x in range(l,r):
    axis[x] += 1
    if axis[x] > k:
      chill = False
  if not chill:
    rR, rL, rInd = heapq.heappop(q)
    rR *= -1
    dummy.append(rInd)
    for x in range(rL,rR):
      axis[x] -= 1
print(len(dummy))
print(' '.join([str(x) for x in dummy]))