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:
[n, k] = list( map( int, input().split(' ')))
arr = []
for index in range( n ):
    inp = list( map( int, input().split(' ')))
    inp.append( index + 1 )
    arr.append( inp )
arr.sort( key=lambda x: x[0] )

pres = []
m = 0
rem = []
for segment in arr:
    pres = list( filter( lambda x: x[1]>= segment[0], pres ) )
    pres.append( segment )
    while( len( pres ) > k ):
        maxi = max( pres, key=lambda x: x[1])
        pres.remove( maxi )
        m+=1
        rem.append( maxi[2] )
print( m )
print( *rem )