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>
#define f(i,n) for(int i=0;i<n;i++)
#define P pair<int,int>
#define F first
#define S second
#define int long long
#define mod 1000000007
#define mmod 998244353
#define all(v) v.begin(), v.end()
#define INF LLONG_MAX/3
#define PI 3.14159265359
using namespace std;
int mypow(int fa, int fb, int fc) {
	if (fb == 0)return 1;
	if (fb % 2)return mypow(fa, fb - 1, fc)*fa%fc;
	int z = mypow(fa, fb / 2, fc);
	return z * z%fc;
}
int k(int fa) {
	int x = 0;
	while (fa) {
		x++;
		fa /= 10;
	}
	return x;
}
int gcd(int fa, int fb) {
	if (fb == 0)return fa;
	return gcd(fb, fa%fb);
}
int x[200000], y[200000];
signed main() {
	int a;
	cin >> a;
	f(i, a) {
		cin >> x[i] >> y[i];
	}
	sort(x, x + a);
	sort(y, y + a);
	if (a % 2) {
		cout << y[a / 2] - x[a / 2] + 1 << endl;
	}
	else {
		cout << y[a / 2] + y[a / 2 - 1] - x[a / 2] - x[a / 2 - 1] + 1 << endl;
	}
}