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;

#define nl printf("\n")
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

using namespace std;

///Output
template<class T>
void print(T y)
{
    if(typeid(y) == typeid(int))printf("%d ",y);
    else if(typeid(y) == typeid(long long))printf("%lld ",y);
    else if(typeid(y) == typeid(long double))printf("%LF ",y);
    else if(typeid(y) == typeid(float))printf("%f ",y);
    else if(typeid(y) == typeid(char))printf("%c ",y);
}

///Input
template<class T>
void read(T &y)
{
    if(typeid(y) == typeid(int))scanf("%d",&y);
    else if(typeid(y) == typeid(long long))scanf("%lld",&y);
    else if(typeid(y) == typeid(long double))scanf("%LF",&y);
    else if(typeid(y) == typeid(float))scanf("%f",&y);
}

int main()
{
    int n;
    read(n);
    int a[n+5],b[n+5];
    for(int i=0;i<n;i++)
    {read(a[i]);read(b[i]);}
    sort(a,a+n);
    sort(b,b+n);
    if(n%2==1)print(b[n/2]-a[n/2]+1),nl;
    else
    {
         double mn,mx,ans;
         mn=(a[n/2-1]+a[n/2])/2.0;
         mx=(b[n/2-1]+b[n/2])/2.0;
         ans=2*(mx-mn)+1;
        cout<<ans<<endl;
    }
}