cf 1029 C

C. Maximal Intersection

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

The intersection of a sequence of segments is such a maximal set of points (not necesserily having integer coordinates) that each point lies within every segment from the sequence. If the resulting set isn't empty, then it always forms some continuous segment. The length of the intersection is the length of the resulting segment or $\mathrm{00 in\; case\; the\; intersection\; is\; an\; empty\; set.}$

For example, the intersection of segments $[1;5][1;5] and [3;10][3;10] is [3;5][3;5] (length 22),\; the\; intersection\; of\; segments [1;5][1;5] and [5;7][5;7] is [5;5][5;5] (length 00)\; and\; the\; intersection\; of\; segments [1;5][1;5] and [6;6][6;6] is\; an\; empty\; set\; (length 00).$

Your task is to remove exactly one segment from the given sequence in such a way that the intersection of the remaining $(n-1)(n-1) segments\; has\; the\; maximal\; possible\; length.$

Input

The first line contains a single integer $\mathrm{nn (2\le n\le 3\cdot 1052\le n\le 3\cdot 105)\; —\; the\; number\; of\; segments\; in\; the\; sequence.}$

Each of the next $\mathrm{nn lines\; contains\; two\; integers lili and riri (0\le li\le ri\le 1090\le li\le ri\le 109)\; —\; the\; description\; of\; the ii-th\; segment.}$

Output

Print a single integer — the maximal possible length of the intersection of $(n-1)(n-1) remaining\; segments\; after\; you\; remove\; exactly\; one\; segment\; from\; the\; sequence.$

Examples

input

Copy

4
1 3
2 6
0 4
3 3

output

Copy

1

input

Copy

5
2 6
1 3
0 4
1 20
0 4

output

Copy

2

input

Copy

3
4 5
1 2
9 20

output

Copy

0

input

Copy

2
3 10
1 5

output

Copy

7

Note

In the first example you should remove the segment $[3;3][3;3],\; the\; intersection\; will\; become [2;3][2;3] (length 11).\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection [3;3][3;3] (length 00).$

In the second example you should remove the segment $[1;3][1;3] or\; segment [2;6][2;6],\; the\; intersection\; will\; become [2;4][2;4] (length 22)\; or [1;3][1;3] (length 22),\; respectively.\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection [2;3][2;3] (length 11).$

In the third example the intersection will become an empty set no matter the segment you remove.

In the fourth example you will get the intersection $[3;10][3;10] (length 77)\; if\; you\; remove\; the\; segment [1;5][1;5] or\; the\; intersection [1;5][1;5] (length 44)\; if\; you\; remove\; the\; segment [3;10][3;10].$

 

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define ll long long 
#define  N 300009
int n,a[N],b[N],c[N],d[N];
map<int,int>mp;
int maxx=-1,minn=1000000009;
bool check(int x,int y)
{
    if(x!=y||x==y&&mp[y]>1)
        return true;
    return false;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a[i],&b[i]);
        c[i]=a[i];d[i]=b[i];
        mp[a[i]]++;mp[b[i]]++;
        maxx=max(maxx,a[i]);
        minn=min(minn,b[i]);
    }
    //cout<<maxx<<" "<<minn<<endl;
    sort(a+1,a+1+n);
    sort(b+1,b+1+n);
    int l=0;
    for(int i=1;i<=n;i++){
        //cout<<maxx<<" "<<minn<<" "<<c[i]<<" "<<d[i]<<endl;
        if(c[i]==maxx&&check(d[i],minn)){
         l=max(l,minn-a[n-1]);
        }
        if(d[i]==minn&&check(c[i],maxx)){
            l=max(l,b[2]-maxx);
        }
        if(c[i]==maxx&&d[i]==minn){//去掉某一个区间，那么区间的左右要同时去掉
            //cout<<l<<endl;
            l=max(l,b[2]-a[n-1]);
            //cout<<b[2]<<" "<<a[n-1]<<endl;
        }
    }
    /*
    2
    1 5
    2 4
    去掉2 4
     2 4要同时去掉
    */
    printf("%d\n",l);
    return 0;        
}