1249_D2. Too Many Segments (hard version)_16238
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 ≤ 2 ⋅ 10^5) — 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 ≤ 2 ⋅ 10^5) — 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
4 6 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 4 5 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 
import heapq

n,k = map(int,input().split())

llis = [ [] for i in range(2 * (10**5) + 1) ]
rlis = [ [] for i in range(2 * (10**5) + 1) ]
flag = [0] * (2 * (10**5) + 1)

for i in range(n):

    l,r = map(int,input().split())
    
    llis[l].append([i+1,r])
    rlis[r].append([i+1,l])

now = 0
rq = []
ans = []
for i in range(2 * (10 ** 5)):
    
    i += 1

    for j in llis[i]:
        now += 1
        ind = j[0]
        r = j[1]
        flag[ind] = 1
        heapq.heappush(rq,[-1 * r,ind])

    for j in rlis[i-1]:
        
        ind = j[0]
        if flag[ind] == 1:
            now -= 1
            flag[ind] = 0

    while now > k:

        nowq = heapq.heappop(rq)
        ind = nowq[1]

        if flag[ind] == 1:
            ans.append(ind)
            now -= 1
            flag[ind] = 0


print (len(ans))
print (" ".join(map(str,ans)))

        
