Problem:
Score : 500 points

Problem Statement

There are N integers X_1, X_2, \cdots, X_N, and we know that A_i \leq X_i \leq B_i.
Find the number of different values that the median of X_1, X_2, \cdots, X_N can take.

Notes

The median of X_1, X_2, \cdots, X_N is defined as follows. Let x_1, x_2, \cdots, x_N be the result of sorting X_1, X_2, \cdots, X_N in ascending order.

If N is odd, the median is x_{(N+1)/2};

if N is even, the median is (x_{N/2} + x_{N/2+1}) / 2.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the answer.

Sample Input 1

2
1 2
2 3

Sample Output 1

3

If X_1 = 1 and X_2 = 2, the median is \frac{3}{2};

if X_1 = 1 and X_2 = 3, the median is 2;

if X_1 = 2 and X_2 = 2, the median is 2;

if X_1 = 2 and X_2 = 3, the median is \frac{5}{2}.

Thus, the median can take three values: \frac{3}{2}, 2, and \frac{5}{2}.

Sample Input 2

3
100 100
10 10000
1 1000000000

Sample Output 2

9991

Solution in Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val AB = Array(N) { Pair(sc.next().toLong(), sc.next().toLong()) }
        val Xs = AB.sortedBy { it.first }
        val Xe = AB.sortedBy { -it.second }
        val ans = if (N % 2 == 1) {
            val mi = (N + 1) / 2 - 1
            Xe[mi].second - Xs[mi].first + 1
        } else {
            val mi1 = N / 2 - 1
            val mi2 = N / 2
            (Xe[mi1].second + Xe[mi2].second) - (Xe[mi1].first + Xe[mi2].first) + 1
        }
        pw.println(ans)
    }
}