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;

int f(const vector<pair<long long, long long>> &ab, size_t i)
{
    long long l = 0;
    long long r = 1e9 + 5;
    while (r - l > 1)
    {
        long long c = (l + r) / 2;
        size_t idx = (lower_bound(ab.begin(), ab.end(), make_pair(c, 0ll)) - ab.begin());
        if (idx > i)
        {
            r = c;
        }
        else
        {
            l = c;
        }
    }
    return l;
}

int g(const vector<pair<long long, long long>> &ba, size_t i)
{
    long long l = 0;
    long long r = 1e9 + 5;
    while (r - l > 1)
    {
        long long c = (l + r) / 2;
        size_t idx = (upper_bound(ba.begin(), ba.end(), make_pair(c, 0ll)) - ba.begin());
        if (idx >= i)
        {
            r = c;
        }
        else
        {
            l = c;
        }
    }
    return l;
}

int main()
{
    size_t N;
    cin >> N;

    vector<long long> A(N);
    vector<long long> B(N);
    for (size_t i = 0; i < N; i++)
    {
        cin >> A[i] >> B[i];
    }
    sort(A.begin(), A.end());
    sort(B.begin(), B.end());

    if (N % 2 == 0)
    {
        long long l2 = A[N / 2] + A[N / 2];
        long long r2 = B[N / 2] + B[N / 2];
        cout << r2 - l2 + 1 << endl;
    }
    else
    {
        long long l = A[N / 2];
        long long r = B[N / 2];
        cout << r - l + 1 << endl;
    }
}