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>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/range/adaptor/reversed.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
typedef long long int ll;

#define PI (acos(-1))
#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define dumparr(x) cerr << #x ; for(auto elem: x){cerr << " " << elem;} cerr << endl;
#define dumppairs(x) cerr << #x ; for(auto elem: x){cerr << "(" << elem.first << " " << elem.second << ") ";} cerr << endl;
#define dumpline cerr << "**********************" << endl;
#define dumpwline(x) cerr << "******** " << #x << "=" << (x) << " ********" << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
#define ARRAY_LENGTH(array) (sizeof(array) / sizeof(array[0]))

int main(void){
    ll n; cin >> n;
    vector <ll> as;
    vector <ll> bs;
    REP(i, n){
        ll a; ll b;
        cin >> a >> b;
        as.push_back(a);
        bs.push_back(b);
    }
    sort(as.begin(), as.end());
    sort(bs.begin(), bs.end());

    ll low = as[n/2];
    if (n % 2 == 0){
        low += as[n/2 - 1];
    }
    ll high = bs[n/2];
    if (n % 2 == 0){
        high += bs[n/2 - 1];
    }

    cout << high - low + 1 << endl;
    return 0;
}