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: Wrong Answer)

package main

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

var sc = bufio.NewScanner(os.Stdin)

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

func intScanner() int {
	n, _ := strconv.Atoi(Scanner())
	return n
}

func floatScanner() float64 {
	n, _ := strconv.ParseFloat(Scanner(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func stringSliceScanner(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scanner()
	}
	return a
}
func intSliceScanner(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = intScanner()
	}
	return a
}

var mod int

func main() {
	mod = 1000000007
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	n := intScanner()
	a := make([]int, n)
	b := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = intScanner() * 2
		b[i] = intScanner() * 2
	}
	am, bm := 0, 0
	if n%2 == 0 {
		am = (a[n/2] + a[n/2-1]) / 2
		bm = (b[n/2] + b[n/2-1]) / 2
	} else {
		am = a[n/2] / 2
		bm = b[n/2] / 2
	}
	fmt.Println(bm - am + 1)
}