codeforces 1000C（差分+pair）

C. Covered Points Count

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every k∈[1..n]k∈[1..n], calculate the number of points with integer coordinates such that the number of segments that cover these points equals kk. A segment with endpoints lili and riri covers point xx if and only if li≤x≤rili≤x≤ri.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of segments.

The next nn lines contain segments. The ii-th line contains a pair of integers li,rili,ri (0≤li≤ri≤10180≤li≤ri≤1018) — the endpoints of the ii-th segment.

Output

Print nn space separated integers cnt1,cnt2,…,cntncnt1,cnt2,…,cntn, where cnticnti is equal to the number of points such that the number of segments that cover these points equals to ii.

Examples

input

Copy

3
0 3
1 3
3 8

output

Copy

6 2 1 

input

Copy

3
1 3
2 4
5 7

output

Copy

5 2 0 

Note

The picture describing the first example:

Points with coordinates [0,4,5,6,7,8][0,4,5,6,7,8] are covered by one segment, points [1,2][1,2] are covered by two segments and point [3][3] is covered by three segments.

The picture describing the second example:

Points [1,4,5,6,7][1,4,5,6,7] are covered by one segment, points [2,3][2,3] are covered by two segments and there are no points covered by three segments.

题意 ：给你个n个线段，每个线段信息为l r，代表覆盖l点到r点。问被覆盖了1-n的点的数量分别有多少。

思路：差分（差分数组的前缀和为原数组）。l，r的范围1e18，map，pair，vector都能过

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define LL long long
#define inf 0x3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl

LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
const int maxn = 1e5*5+100;
using namespace std;
pair<LL,int>p[maxn];
LL ans[maxn];
int main()
{
    int n;
    cin>>n;
    int tail = 0;
    for(int i = 1;i <= n; ++i)
    {
        LL a,b;
        cin>>a>>b;
        p[++tail] = make_pair(a,1);
        p[++tail] = make_pair(b+1,-1); 
    }
    sort(p+1,p+1+tail);
    LL sum = 0;
    for(int i = 1;i <= tail; ++i)
    {
        sum += p[i].second;
        if(p[i].first == p[i+1].first) continue;
        ans[sum] += p[i+1].first - p[i].first;
    }
    for(int i = 1;i <= n; ++i)
        cout<<ans[i]<<" ";
    cout<<endl;

}