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: Runtime Error)

N = int(input())
A = [list(map(int,input().split())) for _ in range(N)]
C = {}
for i in range(N):
    a,b = A[i]
    if a not in C:
        C[a] = 0
    C[a] += 1
    if b not in C:
        C[b] = 0
    C[b] += 1
C = sorted(list(C.keys()))
if N%2==1:
    k = (N+1)//2
    a = C[k-2]
    b = C[-(k-1)]
    print(b-a+1)
else:
    k = N//2
    a = C[k-2]
    a1 = C[k-1]
    b = C[-(k-1)]
    b1 = C[-(k-1)-1]
    print(b+b1-a-a1+1)