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 
import sys

inp = map(int, sys.stdin.read().split())
ii = 0;
n = inp[ii]; ii+=1
k = inp[ii]; ii+=1

segments = [[0, 0] for i in xrange(200)]
intersects = [[] for i in xrange(200)]
counts = [0] * 200

for i in xrange(n):
    segments[i][0] = inp[ii];ii += 1
    segments[i][1] = inp[ii];ii += 1


for i in xrange(n):
    for j in xrange(n):
        if i == j:
            continue
        a = segments[i][0]
        b = segments[i][1]
        c = segments[j][0]
        d = segments[j][1]

        if not(b < c or a > d):
            intersects[i].append(j)
    counts[i] = len(intersects[i])

arr = sorted(enumerate(counts[:n]), key=lambda x: x[1], reverse=True)

res = []
for i in xrange(n):
    ind = arr[i][0]
    c = counts[ind]
    if c >= k:
        for x in intersects[ind]:
            counts[x] -= 1
        counts[ind] = 0
        res.append(ind + 1)


print len(res)
print ' '.join(map(str, res))



