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:

import sys
n,k=map(int,input().split())
t = -1
def input():
    global t
    t += 1
    return data[t]
data=sys.stdin.readlines()
ls=[]
rs=[]
from heapq import heappop,heappush
inf=2*10**5+1
for i in range(n):
    l,r=map(int,input().split())
    ls.append(l)
    rs.append(r)
rs.append(inf)
line=list(range(n))
line2=list(range(n))
line=[(ls[i],i,+1) for i in line]+[(rs[i],-1,i) for i in line2]
line.sort(key=lambda x:(x[0],rs[x[1]]))
#print(line)
count=0
answer=set()
ends=[]
use=[False]*n
for i in range(2*n):
    x,f,o=line[i]
    if f>=0:
        count+=1
        heappush(ends,(-rs[f],f))
    else:
        if count>k:
            d=count-k
            while d>0:
                h,j=ends[0]
                heappop(ends)
                if not use[j]:
                    answer.add(j)
                    d-=1
            count=k



        if o not in answer:
            count-=1
        use[o]=True
print(len(answer))
answer=[i+1 for i in answer]
print(*answer)