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 bisect import bisect_right, bisect_left
N = int(input())
l = -1
r = 10**10
als = [0]*N
bls = [0]*N
for i in range(N):
  a,b = map(int, input().split())
  als[i] = a
  bls[i] = b
if N%2==0:
  l = -1
  r = 10**10
  while l+1<r:
    s = (l+r)//2
    i = bisect_right(als,s)
    if i>=N//2:
      r = s
    else:
      l = s
  A1 = r
  B1 = r+1
  l = -1
  r = 10**10
  while l+1<r:
    s = (l+r)//2
    i = bisect_right(bls,s)
    j = N-i
    if j>N//2:
      l = s
    else:
      r = s
  A2 = r
  B2 = r+1
  ans = B2+A2-A1-B1+1
  print(ans)
else:
  l = -1
  r = 10**10
  while l+1<r:
    s = (l+r)//2
    i = bisect_left(als,s)
    if i>N//2:
      r = s
    else:
      l = s
  A = l
  l = -1
  r = 10**10
  while l+1<r:
    s = (l+r)//2
    i = bisect_left(bls,s)
    j = N-i
    if j>N//2:
      l = s
    else:
      r = s
  B = l
  print(B-A+1)