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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from collections import Counter
import sys

sys.setrecursionlimit(10 ** 6)

mod = 1000000007
inf = int(1e18)

dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]


def inverse(a):
    return pow(a, mod - 2, mod)


def usearch(x, a):
    lft = 0
    rgt = len(a) + 1
    while rgt - lft > 1:
        mid = (rgt + lft) // 2
        if a[mid] <= x:
            lft = mid
        else:
            rgt = mid
    return lft


def main():
    n=int(input())
    s = [tuple(map(int, input().split())) for i in range(n)]
    a = sorted(i[0] for i in s)
    b = sorted(i[1] for i in s)
    lft = a[n//2] + a[-(n//2)-1]
    rgt = b[n//2] + b[-(n//2)-1]
    print((rgt - lft) // 2 * (2 if len(s) % 2 == 0 else 1) + 1)
main()