Educational Codeforces Round 46 (Rated for Div. 2) C - Covered Points Count

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 $n$ 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 $k\in \left[\mathrm{1..}n\right]$, calculate the number of points with integer coordinates such that the number of segments that cover these points equals $k$. A segment with endpoints $l_i$ and $r_i$ covers point $x$ if and only if $l_i\le x\le r_i$.

Input

The first line of the input contains one integer $n$ ($1\le n\le 2\cdot {10}^5$) — the number of segments.

The next $n$ lines contain segments. The $i$-th line contains a pair of integers $l_i,r_i$ ($0\le l_i\le r_i\le {10}^{18}$) — the endpoints of the $i$-th segment.

Output

Print $n$ space separated integers $cn{t}_1,cn{t}_2,\dots ,cn{t}_n$, where $cn{t}_i$ is equal to the number of points such that the number of segments that cover these points equals to $i$.

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 $[0,4,5,6,7,8]$ are covered by one segment, points $[1,2]$ are covered by two segments and point $\left[3\right]$ is covered by three segments.

The picture describing the second example:

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

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 200000 + 10;
struct f
{
	long long head;
	int shu;
}a[2 * MAXN];
long long mark[MAXN];

int com(const f &u , const f &v)
{
	if(u.head == v.head)
		return u.shu > v.shu;
	return u.head < v.head;
}

int main()
{
	int n;
	cin >> n;
	int num = 0;
	for(int i = 1; i <= n * 2; i ++)
	{
		cin >> a[i].head;
		if(i & 1)
			a[i].shu = 1;
		else
			a[i].shu = 0;
	}
	//a[n + 1].head = a[n + 1].tail = 1e18 + 1;
	//cout << endl;
	sort( a + 1, a + 2 * n + 1, com);
	//for(int i = 1 ; i <= 2 * n ; i ++)
	//	cout << a[i].head << " " << a[i].shu << endl;
	//cout << endl;
	int  temp ;
	long long  f = 0;
	for(int i = 1 ; i <= n * 2; i ++)
	{
		//mark[num] += a[i].head - a[i - 1].head;
		if(a[i].shu == 1)
		{
			if(a[i].head - f - 1 >=0)
				mark[num] += a[i].head - f - 1;
			f = a[i].head;
			temp = i;
			int t = num;
			//cout << "!" << " " << t << endl;
			while(a[temp].head == a[temp + 1].head && temp + 1 <= n * 2)
			{
				//cout << a[temp + 1].shu << endl;


				//t ++;
				if(a[temp + 1].shu > 0)
				{
					num ++;
					t ++;
					//cout << "!" << endl;
				}
					//num ++;
				else
					num --;
				temp ++;
			}
			//cout << t << endl;
			num ++;
			t ++;
			//cout << num << " " << t << endl;
			mark[t] ++;
			i += (temp - i);
			//for(int j = 1; j <= 3 ; j ++)
			//	cout << mark[j] << " ";
			//cout << endl;
		}
		else
		{
			if(a[i].head - f - 1 >= 0)
				mark[num] += a[i].head - f - 1;
			f = a[i].head;
			int t = num ;
			temp = i;
			while(a[temp].head == a[temp + 1].head && temp + 1 <= n * 2)
			{
				if(a[temp + 1].shu > 0)
				{
					num ++;
					t ++;
				}
				//	num ++;
				else
					num --;
				temp ++;
			}
			//cout << num << " " << t << endl;
			mark[t] ++ ;
			num --;

			//for(int j = 1 ; j <= 3; j ++)
			//	cout << mark[j] << " ";
			//cout << endl;
			i += (temp - i) ;
		}
	}
	cout << mark[1];
	for(int i = 2; i <= n ; i ++)
		cout << " " << mark[i];
	return 0;
}