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 ≤ 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 

Solution (in PYTHON3):

Solution:
import sys
from heapq import *
n, k = map(int, input().split())
N = int(2e5+2)
ll = [[] for _ in range(N)]
rr = [[] for _ in range(N)]
vis = [0] * N

for i in range(n):
    l, r = map(int, sys.stdin.readline().split())
    ll[l].append((-r, i+1))
    rr[r].append(i+1)

q = []
ans = []
size = 0

for i in range(1, N):
    for j in ll[i]:
        heappush(q, j)
        size += 1

    while size > k:
        cur = heappop(q)
        while vis[cur[1]]:
            cur = heappop(q)
        vis[cur[1]] = 1
        ans.append(cur[1])
        size -= 1
    for j in rr[i]:
        if not vis[j]:
            vis[j] = 1
            size -= 1

print(len(ans))
print(*ans)