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<math.h>
#include<algorithm>
#include<stdint.h>
#include<vector>
#include<deque>
#include<stack>
#include<functional>
#include<string>
#include<cstring>
#include<time.h>
#include<array>
#include<iomanip>
#include<list>
#include<set>
#include<unordered_map>
#include<unordered_set>
#include<random>
#include<bitset>
#include <queue>
#include <map>


using namespace std;
using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
using ldouble = long double;


//BDD,ZDD,kdtree, bdtree,bicalc, bolonoy, doloney, tree, chinesemod,segmenttree,daikusutora, saidairyuu, 2bugurahu, heirokenshutu, topologicalsort, kyourenketuseibun

#define REP(i,a,b) for(int i = a; i < b; ++i)
#define PRI(s) cout << s << endl
#define PRIY PRI("Yes")
#define PRIN PRI("No")




int main() {
	vector<ll> a, b;
	int N;
	cin >> N;
	REP(i, 0, N) {
		ll a_, b_;
		cin >> a_ >> b_;
		a.push_back(a_);
		b.push_back(b_);
	}
	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	if (N % 2 == 1) {
		PRI(b[N / 2] - a[N / 2] + 1);
	}
	else {
		PRI((b[N / 2 - 1] + b[N / 2]) - (a[N / 2 - 1] + a[N / 2]) + 1);
	}
	return 0;
}