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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#import<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long LL;
LL i,j,k,m,n,s,a[200040],b[200040],c[200040],d[200040];
main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cout<<fixed<<setprecision(12);
	cin>>n;
	for(;i<n;i++)
	{
		cin>>a[i]>>b[i];
		c[i]=a[i];
		d[i]=b[i];
	}
	sort(c,c+n);
	sort(d,d+n);
	if(n%2)
	{
		cout<<d[n/2]-c[n/2]+1;
	}
	else
	{
		cout<<d[n/2-1]+d[n/2]-c[n/2-1]-c[n/2]+1;
	}
}