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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
	int N;
	cin >> N;
	vector< pair<int, int>> AB(N);
	for (int i = 0; i < N; i++) cin >> AB[i].first >> AB[i].second;

	if (N % 2 == 0) {
		sort(AB.begin(), AB.end());
		double Amin = AB[N / 2 - 1].first;
		double Amax = min(AB[N / 2 - 1].second, AB[N / 2].second);
		sort(AB.begin(), AB.end(), [](auto& a, auto& b) {return a.second > b.second; });
		double Bmin = max(AB[N / 2].first, AB[N / 2 - 1].first);
		double Bmax = AB[N / 2 - 1].second;
		cout << ((Amax + Bmax) / 2 - (Amin + Bmin) / 2) * 2 + 1 << endl;
	}
	else {
		sort(AB.begin(), AB.end());
		double Amin = AB[(N - 1) / 2].first;
		double Amax = min(AB[(N - 1) / 2].second, AB[(N + 1) / 2].second);
		sort(AB.begin(), AB.end(), [](auto& a, auto& b) {return a.second > b.second; });
		double Bmin = max(AB[(N - 1) / 2].first, AB[(N + 1) / 2].first);
		double Bmax = AB[(N - 1) / 2].second;
		if (Amax < Bmin) cout << (Amax - Amin) + 1 + (Bmax - Bmin) + 1;
		else cout << Bmax - Amin + 1;
	}

}