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.lang.Math.max

fun main() = Thread(null, ::abc169e, "(^o^)", 128 * 1024 * 1024).apply { start() }.join()

@OptIn(ExperimentalStdlibApi::class)
fun abc169e() {
    fun LongRange.getRange() = this.last - this.first + 1

    val n = readLine()!!.toInt()
    val abList = Array(n) {
        val (a, b) = readLine()!!.split(' ').map { it.toLong() }
        a..b
    }.sortedBy { it.first }

    val answer = if (n % 2 == 0) {
        val leftDown = abList.take(n / 2).map { it.first }.max()!!
        val rightDown = abList.take(n / 2 + 1).map { it.first }.max()!!
        val leftUp = abList.takeLast(n / 2 + 1).map { it.last }.min()!!
        val rightUp = abList.takeLast(n / 2).map { it.last }.min()!!

        max((leftUp + rightUp) - (leftDown + rightDown) + 1, 1)
    } else {
        abList[n / 2].getRange()
    }

    println(answer)
}