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)

n = int(input())
num = []


for i in range(n):
    a, b = map(int, input().split())
    num.append((a, b))
l = 0
r = 10**9+1
while r - l > 1:
    m = l + (r-l)//2
    count = 0
    for i in range(n):
        if num[i][1] >= m:
            count += 1

    if count >= (n+1)//2:
        l = m
    else:
        r = m

right = l


l = -1
r = 10**9+1
while r - l > 1:
    m = l + (r-l)//2
    count = 0
    for i in range(n):
        if m >= num[i][0]:
            count += 1

    if count >= (n+1)//2:
        r = m
    else:
        l = m
left = r
if n % 2 == 1:

    print(right-left+1)
else:
    ans = (right-left)*2-1

    c = 0
    for i in range(n):
        if num[i][0] <= left:
            c += 1
    if c >= n//2 + 1:
        ans += 1

    c = 0
    for i in range(n):
        if num[i][1] >= right:
            c += 1
    if c >= n//2 + 1:
        ans += 1
    print(ans)