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.io.PrintWriter

fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ")
fun nextIntList() = nextList().map(String::toInt)
fun nextLongList() = nextList().map(String::toLong)
fun nextDoubleList() = nextList().map(String::toDouble)

val pw = PrintWriter(System.out)
fun print(s: Any = "") = pw.print(s)
fun println(s: Any = "") = pw.println(s)

fun main(args: Array<String>) {
    main()
    pw.flush()
}

fun main() {
    val N = nextInt()
    val A = mutableListOf<Long>()
    val B = mutableListOf<Long>()
    repeat(N) {
        val (a, b) = nextLongList()
        A.add(a)
        B.add(b)
    }
    A.sort()
    B.sort()
    if (N % 2 == 1) {
        val l = A[N / 2]
        val r = B[N / 2]
        println(r - l + 1)
    } else {
        val ll = A[N / 2] + A[N / 2 - 1]
        val rr = B[N / 2] + B[N / 2 - 1]
        println(rr - ll + 1)
    }
}