D2. Too Many Segments (hard version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions is constraints.

You are given nn segments on the coordinate axis OXOX. Segments can intersect, lie inside each other and even coincide. The ii-th segment is [li;ri][li;ri] (li≤rili≤ri) and it covers all integer points jj such that li≤j≤rili≤j≤ri.

The integer point is called bad if it is covered by strictly more than kk 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 nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next nn lines contain segments. The ii-th line contains two integers lili and riri (1≤li≤ri≤2⋅1051≤li≤ri≤2⋅105) — the endpoints of the ii-th segment.

Output

In the first line print one integer mm (0≤m≤n0≤m≤n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print mm distinct integers p1,p2,…,pmp1,p2,…,pm (1≤pi≤n1≤pi≤n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

input

Copy

output

Copy

3
4 6 7

input

Copy

output

Copy

3
1 4 5

input

Copy

output

Copy

4
1 3 5 6

直接贪心即可

从前往后找发现每个点大于k那么就删掉覆盖该点所有线段上的结束点最靠后的的线段即可

考虑时间复杂度

我们只需要维护一个set集合

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int s;
    int e;
    int id;
}a[200005];
int cmp(node p,node q)
{
    return p.s<q.s;
}
vector<int> ans;
set<pair<int,int> > st;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    int maxn,minn;
    pair<int,int>tmp;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].s,&a[i].e);
        maxn=max(maxn,a[i].e);
        minn=min(minn,a[i].s);
        a[i].id=i;
    }
    sort(a+1,a+1+n,cmp);
    int pos=1;
    for(int i=minn;i<=maxn;i++)
    {
        while(pos<=n&&a[pos].s<=i)
        {
            st.insert(make_pair(a[pos].e,a[pos].id));
            pos++;
        }
        while(st.size()&& st.begin()->first<i)
        {
            st.erase(st.begin());
        }
        while(st.size()>k)
        {
            tmp=*(--st.end());
            ans.push_back(tmp.second);
            st.erase(tmp);
        }
    }
    sort(ans.begin(),ans.end());
    printf("%d
",ans.size());
    for(auto v:ans)
    {
        printf("%d ",v);
    }
    printf("
");
}