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>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T = int> void print(vector<T> v){for(auto a:v)cout<<a<<" ";cout<<endl;}

/*
*/

// struct d{ll a,b;};
// bool operator<(const d &l, const d &r){
// 	if(l.a<r.a)return true;
// 	if(l.a==r.a && l.b>r.b)return true;
// 	return false;
// }
// d p[200010];

ll n,a[200010],b[200010];

int main(void){
	cin>>n;
	rep(i,n){
		cin>>a[i]>>b[i];
	}
	sort(a,a+n);
	sort(b,b+n);
	if(n%2==1){
		cout<<b[n/2]-a[n/2]+1<<endl;
	}else{
		cout<<(b[n/2]+b[n/2-1])-(a[n/2]+a[n/2-1])+1<<endl;
	}
	return 0;
}