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 Go (1.14.1) (judge verdict: Time Limit Exceeded)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)
var out = bufio.NewWriter(os.Stdout)

func next() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	a, _ := strconv.Atoi(next())
	return a
}

func comp(left, right int) bool {
	return left > right
}

func partition(a []int, l int, r int) int {
	pivot := a[r]
	i := l
	for j := l; j < r; j++ {
		if comp(a[j], pivot) {
			a[i], a[j] = a[j], a[i]
			i += 1
		}
	}
	a[i], a[r] = a[r], a[i]
	return i
}

func quicksort(a []int, l int, r int) {
	if l < r {
		p := partition(a, l, r)
		quicksort(a, l, p-1)
		quicksort(a, p+1, r)
	}
}

func main() {
	buf := make([]byte, 1024*1024)
	sc.Buffer(buf, bufio.MaxScanTokenSize)
	sc.Split(bufio.ScanWords)
	defer out.Flush() // !!!!caution!!!! you must use Fprint(out, ) not Print()
	/* --- code --- */
	N := nextInt()
	A := make([]int, N)
	B := make([]int, N)
	for i := 0; i < N; i++ {
		A[i], B[i] = nextInt(), nextInt()
	}

	quicksort(A, 0, N-1)
	quicksort(B, 0, N-1)

	if N%2 == 0 {
		med_min_mult2 := A[N/2-1] + A[N/2]
		med_max_mult2 := B[N/2-1] + B[N/2]
		fmt.Println(med_max_mult2 - med_min_mult2 + 1)
	} else {
		med_min := A[N/2]
		med_max := B[N/2]
		fmt.Println(med_max - med_min + 1)
	}

}