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 <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define rep(i, j, n) for(int i = (j); i < (n); ++i)

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);


	int n; cin >> n;
	vector<int> a(n), b(n);
	rep(i, 0, n)
		cin >> a[i] >> b[i];
	
	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	if (n & 1) {
		cout << b[n / 2] - a[n / 2] + 1 << endl;
	}
	else {
		int l = a[n / 2 - 1] + a[n / 2];
		int r = b[n / 2 - 1] + b[n / 2];
		cout << r - l + 1 << endl;
	}
	return 0;
}