Nested Segments CodeForces

D. Nested Segments

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.

Each of the next n lines contains two integers li and ri ( - 109 ≤ li < ri ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.

Output

Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.

Examples

input

41 82 34 75 6

output

3010

input

33 41 52 6

output

011

给定n条线段，问每条线段包含了几条线段，先根据左端点值从大到小排序，相等按照右端点值从小到大，这样排序后前面的区间相对与后面的区间左端点可以满足条件，只需求前面的线段里有几个右端点也满足的即可，这一步可以用树状数组维护，不过这个题的右端点值范围较大，所以要先将右端点值离散化处理，在用树状数组维护

#include<bits/stdc++.h>using namespace std;struct node{int l,r;int index;}num[200005];int f[200005],ans[200005],n;int sum(int x){int sum=0;while(x>0){sum+=f[x];x-=x&-x;}return sum;}int add(int t,int x){while(t<=n){f[t]+=x;t+=t&-t;}}int cmp1(node u,node v){if(u.l==v.l){return u.r<v.r;}return u.l>v.l;}int cmp(node u,node v){return u.r<v.r;}map<int,int>p;int main(){int i;cin>>n;for(i=0;i<n;i++){scanf("%d %d",&num[i].l,&num[i].r);num[i].index=i;}int cnt=1;sort(num,num+n,cmp);for(i=0;i<n;i++){if(!p.count(num[i].r)){p[num[i].r]=cnt++;}}sort(num,num+n,cmp1);for(i=0;i<n;i++){ans[num[i].index]=sum(p[num[i].r]);add(p[num[i].r],1);}for(i=0;i<n;i++){printf("%d\n",ans[i]);}return 0;}