CodeForces C. Maximal Intersection

http://codeforces.com/contest/1029/problem/C

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

代码：

#include <bits/stdc++.h>
using namespace std;

#define inf 0x3f3f3f3f
const int maxn = 300010 + 10;
int N;

struct Node {
    int l;
    int r;
}S[maxn], Q[maxn], A[maxn];

int main() {
    scanf("%d", &N);
    S[0].r = inf, S[0].l = -inf;
    for(int i = 1; i <= N; i ++) {
        scanf("%d%d", &A[i].l, &A[i].r);
        S[i].l = max(S[i - 1].l, A[i].l);
        S[i].r = min(S[i - 1].r, A[i].r);
    }

    Q[N + 1].r = inf, Q[N + 1].l = -inf;
    for(int i = N; i >= 1; i --) {
        Q[i].l = max(A[i].l, Q[i + 1].l);
        Q[i].r = min(A[i].r, Q[i + 1].r);
    }

    int ans = 0;
    for(int i = 1; i <= N; i ++) {
        ans = max(ans, (min(Q[i + 1].r, S[i - 1].r) - max(Q[i + 1].l, S[i - 1].l)));
    }
    printf("%d
", ans);
    return 0;
}