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 Python (3.8.2) (judge verdict: Wrong Answer)

import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub, or_

sys.setrecursionlimit(100000)
input = sys.stdin.readline
INF = 2**62-1

def read_int():
    return int(input())

def read_int_n():
    return list(map(int, input().split()))

def read_float():
    return float(input())

def read_float_n():
    return list(map(float, input().split()))

def read_str():
    return input().strip()

def read_str_n():
    return list(map(str, input().split()))

def error_print(*args):
    print(*args, file=sys.stderr)


def main():
    n = read_int()
    al = []
    bl = []
    for i in range(n):
        a, b = read_int_n()
        al.append(a)
        bl.append(b)
    al.sort()
    al.reverse()
    bl.sort()
    out = (math.ceil(n/2) - 1)
    
    cl = al[n - 1 - out]
    ch = bl[n - 1 - out]

    if n % 2 == 1:
        print(ch - cl + 1)
    else:
        c = ch - cl + 1
        print(math.floor(c * (c-1) / 2))


if __name__ == '__main__':
    main()