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 pb push_back
#define mp make_pair
#define mod 1000000007
#define int long long int
#define double long double
#define inf (int)(1e15)
#define all(x) (x).begin(), (x).end()
#define pair pair<int, int>
typedef vector<int> vi;   // Vector of long long
typedef vector<vi> vvi;   // Vector of vi
typedef vector<pair> vii; // Vector of pairs
typedef vector<vii> vvii; // Vector of Vector of pairs
typedef vector<bool> vb;  // Vector of bool
#define pq priority_queue // Max heap (To convert to min heap, use negative sign before every value)
#define ff first          // For pairs
#define ss second
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, -1, 0, 1};
const int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1}, dy8[] = {1, -1, 1, 0, -1, 1, -0, -1};
#define testcases(t) \
    int(t);          \
    cin >> (t);      \
    while ((t)--)
signed main()
{
    int n;
    cin >> n;
    vi a(n + 1);
    vi b(n + 1);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i] >> b[i];
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    if (n % 2)
    {
        int x = a[n / 2], y = b[n / 2];
        cout << (y - x + 1) << endl;
    }
    else
    {
        int x1 = a[n / 2], x2 = a[n / 2 + 1], y1 = b[n / 2], y2 = b[n / 2 + 1];
        cout << (y2 + y2 - x2 - x2 + 1) << endl;
    }
    return 0;
}