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: Runtime Error)

import sys

# import re
import math
import collections
# import decimal
import bisect
import itertools
import fractions
# import functools
import copy
import heapq
import decimal
# import statistics
import queue

sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7

ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))


# ===CODE===


def main():
    n = ni()
    a = []
    b = []
    d = []
    for _ in range(n):
        ai, bi = ns()
        a.append(ai)
        b.append(bi)
        d.append((ai, bi))

    a.sort()
    b.sort()

    odd = n % 2 == 1
    even = n % 2 == 0
    min_choice = a[n // 2 - even]
    max_choice = b[n // 2]

    kouho = []
    for ai, bi in d:
        tmp = range(max(ai, min_choice), min(bi, max_choice))
        if len(tmp) > 0:
            kouho.append(tmp)

    dic_even = dict()
    dic_odd = dict()

    for k in kouho:
        for t in k:
            dic_even[t] = 0

    for i in range(len(kouho)):
        for t in kouho[i]:
            for j in range(i + 1, len(kouho)):
                for l in kouho[j]:
                    dic_odd[(t + l) / 2] = 0
    
    e = set(dic_even.keys())
    o = set(dic_odd.keys())
    
    ans = e+o
    print(len(ans))


if __name__ == '__main__':
    main()