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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <numeric>
#include <utility>
#include <string>

using namespace std;

using PII = pair<int, int>;

int main() {
	int n;
	cin >> n;

	int a, b;
	vector<PII> ab1;
	for (int i = 0; i < n; i++) {
		cin >> a >> b;
		ab1.emplace_back(a, b);
	}

	sort(ab1.begin(), ab1.end());
	auto ab2 = ab1;
	sort(ab2.begin(), ab2.end(),
		[](const PII &x, const PII &y) {
			if (x.second == y.second) {
				return x.first > y.first;
			} else {
				return x.second > y.second;
			}
		});

	// printf("----\n");
	// for (auto p: ab1) {
	// 	printf("%10d %10d\n", p.first, p.second);
	// }
	// printf("----\n");
	// for (auto p: ab2) {
	// 	printf("%10d %10d\n", p.first, p.second);
	// }

	int lower1, upper1, lower2, upper2;
	if (n % 2 == 0) {
		int mid2 = n / 2;
		int mid1 = mid2 - 1;

		lower1 = ab1[mid1].first + ab1[mid2].first;
		upper1 = ab1[mid1].second + ab1[mid2].second;

		lower2 = ab2[mid1].first + ab2[mid2].first;
		upper2 = ab2[mid1].second + ab2[mid2].second;
	} else {
		// 奇数個
		if (n == 1) {
			lower1 = lower2 = ab1[0].first;
			upper1 = upper2 = ab1[0].second;
		} else {

			int mid = n / 2;
			lower1 = ab1[mid].first;
			upper1 = min(ab1[mid].second, ab1[mid + 1].second);

			upper2 = ab2[mid].second;
			lower2 = max(ab2[mid].first, ab2[mid + 1].second);

		}
	}

	int ans;
	if (upper1 < lower2 || upper2 < lower1) {
		ans = (upper1 - lower1 + 1) + (upper2 - lower2 + 1);
	} else {
		int lower = min(lower1, lower2);
		int upper = max(upper1, upper2);
		ans = upper - lower + 1;
	}

	cout << ans << endl;
	return 0;
}