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<vector>
#include<map>
#include<set>
#include<cassert>
#include<cassert>
#include<unordered_map>
#include<unordered_set>
#include<functional>
#include<queue>
#include<stack>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<sstream>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<numeric>
#include<random>
#include<chrono>
#include<bitset>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define pb push_back
#define xx first
#define yy second
#define sz(x) (int)(x).size()
#define gc getchar
#define IO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mp make_pair

#ifndef ONLINE_JUDGE
#  define LOG(x) (cerr << #x << " = " << (x) << endl)
#else
#  define LOG(x) ((void)0)
#endif

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const double PI=acos(-1);
const ll INF = 1LL<<62;
const ll MINF = -(1LL<<62);

template<typename T> T getint() {
	T val=0;
	char c;
	
	bool neg=false;
	while((c=gc()) && !(c>='0' && c<='9')) {
		neg|=c=='-';
	}

	do {
		val=(val*10)+c-'0';
	} while((c=gc()) && (c>='0' && c<='9'));

	return val*(neg?-1:1);
}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); uniform_int_distribution<int>(0, n-1)(rng)


int main() {
	IO;
	int n;
	cin>>n;
	vector<int> a(n),b(n);
	for(int i=0;i<n;++i) {
		cin>>a[i]>>b[i];
	}
	sort(all(a));
	sort(all(b));
	if(n&1) {
		cout<<(b[n/2]-a[n/2]+1)<<"\n";
	}else {
		cout<<(b[n/2]+b[n/2-1])-(a[n/2]+a[n/2-1])+1<<"\n";
	}
	return 0;
}