1249_D1. Too Many Segments (easy version)
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 
```python
import sys
from heapq import *
#sys.stdin = open('in', 'r')
#n = int(input())
n,k = map(int, input().split())
seg = []
for i in range(n):
    l, r = map(int, input().split())
    seg.append((l, r, i+1))
seg.sort()
c = 0
res = []
i = 0
hmin = []
hmax = []
rem = set()
while i < n:
    l,r,si = seg[i]
    while i < n and seg[i][0] == l:
        _,r,si = seg[i]
        heappush(hmax, (-r, si))
        heappush(hmin, (r, si))
        i += 1
        c += 1
    while len(hmin) > 0 and hmin[0][0] < l:
        mr,ind = heappop(hmin)
        if ind not in rem:
            c -= 1
    while c > k:
        mr,ind = heappop(hmax)
        res.append(ind)
        rem.add(ind)
        c -= 1
print(len(res))
if len(res) > 0:
    print(' '.join(map(str, res)))
        
        





#sys.stdout.write('YES\n')
#sys.stdout.write(f'{res}\n')
#sys.stdout.write(f'{y1} {x1} {y2} {x2}\n')

```