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 operator import itemgetter
import sys
input = sys.stdin.readline


n, m = map(int, input().split())
info = [list(map(int, input().split())) + [i] for i in range(n)]
info = sorted(info, key = itemgetter(1))
max_num = info[-1][1]

N = max_num
INF = 0
LV = (N-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [0]*(2*N0)

def gindex(l, r):
    L = (l + N0) >> 1; R = (r + N0) >> 1
    lc = 0 if l & 1 else (L & -L).bit_length()
    rc = 0 if r & 1 else (R & -R).bit_length()
    for i in range(LV):
        if rc <= i:
            yield R
        if L < R and lc <= i:
            yield L
        L >>= 1; R >>= 1

def propagates(*ids):
    for i in reversed(ids):
        v = lazy[i-1]
        if not v:
            continue
        lazy[2*i-1] += v; lazy[2*i] += v
        data[2*i-1] += v; data[2*i] += v
        lazy[i-1] = 0

def update(l, r, x):
    *ids, = gindex(l, r)
    propagates(*ids)

    L = N0 + l; R = N0 + r
    while L < R:
        if R & 1:
            R -= 1
            lazy[R-1] += x; data[R-1] += x
        if L & 1:
            lazy[L-1] += x; data[L-1] += x
            L += 1
        L >>= 1; R >>= 1
    for i in ids:
        data[i-1] = max(data[2*i-1], data[2*i])

def query(l, r):
    propagates(*gindex(l, r))
    L = N0 + l; R = N0 + r

    s = INF
    while L < R:
        if R & 1:
            R -= 1
            s = max(s, data[R-1])
        if L & 1:
            s = max(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s

ans = []
for i in range(n):
    l, r, j = info[i]
    r += 1
    if query(l, r) < m:
        update(l, r, 1)
    else:
        ans.append(j+1)
print(len(ans))
print(*ans)