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: Accepted)

import java.lang.ArithmeticException

fun str() = readLine()!!
fun int() = str().toInt()
fun long() = str().toLong()

fun read(d:String = " ") = str().split(d)
fun <T> list(d:String = " ",f:String.()->T) = read(d).map {f(it)}
fun list() = list {toInt()}.toIntArray()

fun <T> T.print() = this.also(::println)

fun main() {
	abc000X()
}

fun abc000X() {
	val n = int()
	val a = LongArray(n) {0}
	val b = LongArray(n) {0}
	for(i in a.indices) {
		val (t,u) = list {toLong()}
		a[i] = t
		b[i] = u
	}
	a.sort()
	b.sort()
//	System.err.println(a.joinToString())
//	System.err.println(b.joinToString())

	if(n%2==1) {
		println(b[n/2]-a[n/2]+1)
	} else {
		val l = a[n/2-1]+a[n/2]
		val h = b[n/2-1]+b[n/2]
		println(h-l+1)
	}
}