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)

n=int(input())
AB=[]
for i in range(n):
    a,b=map(int,input().split())
    AB.append((a,b))

AB.sort()
ab=AB[:]
ab.sort(key=lambda x:x[1])

if n%2==0:
    left=AB[n//2-1][0]
    right=ab[n//2][1]
    cnt=0
    cnt2=0
    
    nl=AB[0][0]
    nr=AB[0][1]
    for i in range(1,n):
        now=AB[i]
        if now[0]<=nr:
            nr=max(nr,now[1])
            #print(cnt,cnt2)
            cnt2+=cnt*(min(nr,right)-max(now[0],left))-((min(nr,right)-max(now[0],left))*(min(nr,right)-max(now[0],left)-1))//2
            #print(cnt,cnt2)
        else:
            #print(cnt,cnt2)
            cnt2+=cnt*(min(nr,right)-max(now[0],left))-((min(nr,right)-max(now[0],left))*(min(nr,right)-max(now[0],left)-1))//2
            #print(cnt,cnt2)
            cnt+=max(min(nr,right)-max(nl,left)+1,0)
            #print(cnt,cnt2)
            nl=now[0]
            nr=now[1]
    #print(cnt,cnt2)
    cnt2+=cnt*(min(nr,right)-max(now[0],left))-((min(nr,right)-max(now[0],left))*(min(nr,right)-max(now[0],left)-1))//2
    #print(cnt,cnt2)
    cnt+=max(min(nr,right)-max(nl,left)+1,0)
    #print(cnt,cnt2)
    print(cnt2)


else:
    left=AB[n//2][0]
    right=ab[n//2][1]
    cnt=0
    
    nl=AB[0][0]
    nr=AB[0][1]
    for i in range(1,n):
        now=AB[i]
        if now[0]<=nr:
            nr=max(nr,now[1])
        else:
            cnt+=max(min(nr,right)-max(nl,left)+1,0)
            nl=now[0]
            nr=now[1]
    cnt+=max(min(nr,right)-max(nl,left)+1,0)
    print(cnt)