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)

#include <bits/stdc++.h>
using namespace std;

int main(){
	long long n;
	cin >> n;
	vector<int> a(n), b(n);
	for(int i = 0 ; i < n; i++){
		cin >> a[i] >> b[i];
	}
	sort(a.begin(),a.end());
	sort(b.begin(),b.end());
	long long low, high;
	long long mid1, mid2;
	if(n % 2 == 0){
		mid1 = (n-1)/2;
		mid2 = n/2;
		low = min(a[mid1],b[mid1])+min(a[mid2],b[mid2]);
		high = max(a[mid1],b[mid1]) + max(a[mid2],b[mid2]);
	}else{
		mid1 = n / 2;
		low = min(a[mid1],b[mid1]);
		high = max(a[mid1],b[mid1]);
	}
	cout << high - low + 1 << "\n";
	return 0;
}