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)

// abc169_e
#include <bits/stdc++.h>
#ifdef LOCAL
#include "../cxx-prettyprint/prettyprint.hpp"
#endif
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))
#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))
#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))
#define ALL(x) x.begin(), x.end()

#define INF (ll)(1e9)
#define MOD (1000000007)

#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }
#define print_line(vec, n) {for(int i=0;i<(n-1);i++) cout << (vec)[i] << " "; cout << (vec)[(n)-1] << endl;}
template<class T> void print(const T& x){cout << x << endl;}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }
struct PreMain {PreMain(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} premain;

int main() {
#ifdef LOCAL
    ifstream in("../arg.txt"); cin.rdbuf(in.rdbuf());
#endif

    int N;
    cin >> N;
    vector<P> AB(N);
    REP(i, N) cin >> AB[i].first >> AB[i].second;

    ll ans;
    if (N % 2 == 0){
        sort(ALL(AB));
        ll mn = AB[N/2-1].first;
        ll mn2 = AB[N/2].first;

        sort(ALL(AB), [](P& a, P& b){
            return a.second > b.second;
        });
        ll mx = AB[N/2-1].second;
        ll mx2 = AB[N/2].second;

        ans = ((mx+mx2) - (mn+mn2) + 1);
    } else {

        sort(ALL(AB));
        ll mn = AB[N/2].first;

        sort(ALL(AB), [](P& a, P& b){
            return a.second > b.second;
        });
        ll mx = AB[N/2].second;

        ans = (mx - mn + 1);

    }

    print(ans);


    return 0;
}