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 rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,b,a) for(int i=b;i>=a;i--)
#define fori(a) for(auto i : a )
#define all(a) begin(a), end(a)
#define set(a,b) memset(a,b,sizeof(a))
#define sz(a) a.size()
#define pi 3.14159
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PF push_front //deque
#define mp make_pair
#define pq priority_queue
#define mod 1000000007
#define f first
#define s second
#define pii pair< int, int >
#define vi vector<int>
#define vpii vector<pii>
#define debug(v) for(auto i:v) cout<<i<<" ";
#define tc int t; cin >> t; while(t--)

using namespace std;
string repeat(string s, int n) {
    string s1 = "";
    for (int i=0; i<n;i++)
        s1+=s;
    return s1;
}
string getString(char x) {
    string s(1, x);
    return s;
}

void optimizeIO(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
int gcd(int a, int b){
    if (a == 0)  return b;
    return gcd(b % a, a);
}

void solve(){
  ll n;
  cin>>n;
  vector<ll> v1(n),v2(n);
  rep(i,0,n) cin>>v1[i]>>v2[i];
  sort(all(v1));
  sort(all(v2));
  if(n%2) cout<<v2[n/2]-v1[n/2]+1LL<<endl;
  else{
    double x = (v1[n/2LL]+v1[(n)/2LL-1]), zy= (v2[n/2]+v2[(n)/2LL-1]);
    double z=(zy-x)+1;
    cout<<(ll)z<<endl;
  }
}
int main(){
    optimizeIO();
    // tc
    {
      solve();
    }
}