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 rgi register int
#define ll long long
class fastin{
	private: int _ch,_f;
	public: inline fastin& operator>>(char&c) {c=getchar(); return *this;}
		template<typename _Tp> inline fastin& operator>>(_Tp&_x){
		_x=0; while(!isdigit(_ch)) _f|=(_ch==45),_ch=getchar();
		while(isdigit(_ch)) _x=(_x<<1)+(_x<<3)+(_ch^48),_ch=getchar();
		_f&&(_x=-_x,_f=0); return *this;}
	fastin() {_ch=_f=0;}
}fin;
class fastout{
#define endl '\n'
	private: int _num[32],_head;
	public: inline fastout& operator<<(char c) {putchar(c); return *this;}
		template<typename _Tp> inline fastout& operator<<(_Tp _x){
		_Tp _k; if(_x==0) {putchar('0'); return *this;} if(_x<0) putchar('-'),_x=-_x;
		while(_x>0) _k=_x/10,++_head,_num[_head]=(_x-(_k<<1)-(_k<<3))^48,_x=_k;
		while(_head>0) putchar(_num[_head]),--_head; return *this;}
	fastout() {_head=0;}
}fout;
// ----------------------------
#define int ll
// using namespace std;
const int maxn=200004;
const int mod=998244353,inf=1000000007;

// struct query{int l,int r}q[maxn];
// inline bool cmp(query a,query b) {return a.l!=b.l?a.l<b.l:a.r<b.r;}
int a[maxn],b[maxn];
signed main()
{
	ll n;
	fin>>n;
	for(rgi i=1;i<=n;++i) fin>>a[i]>>b[i];
	std::sort(a+1,a+1+n),std::sort(b+1,b+1+n);
	if(n&1) fout<<(b[n/2+1]-a[n/2+1]+1);
	else fout<<(b[n/2]+b[n/2+1]-a[n/2]-a[n/2+1]+1);
	return 0;
}
// ----------------------------
// by imzzy