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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long int LLI;

void solve()
{
	LLI n;
	cin>>n;
	vector<LLI>A(n),B(n);
	for(LLI i=0;i<n;++i) cin>>A[i]>>B[i];
	sort(A.begin(),A.end());
	sort(B.begin(),B.end());
	LLI median_max,median_min,idx;
	if(n & 1){
		idx=n/2;
		median_min=A[idx];
		median_max=B[idx];
	}
	else{
		idx=n/2;
		median_min=(A[idx]+A[idx-1])/2;
		median_max=(B[idx]+B[idx-1])/2;
	}
	cout<<median_max-median_min+1;
	return;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	solve();
	return 0;
}