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>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

typedef long long ll;
const int N = 2e5 + 10;
int n;
ll a[N], b[N], ans;

int main() {
    cin >> n;
    rep(i, 1, n) scanf("%lld%lld", &a[i], &b[i]);
    if (n & 1) return printf("%lld\n", b[(n + 1) / 2] - a[(n + 1) / 2] + 1), 0;
    ll x = n / 2, y = x + 1;
    ll l = b[x] - a[x] + 1, r = b[y] - a[y] + 1;
    /*if (l < r) {
        ans = r + (l - 1);
    } else if (l > r) {
        ans = l + (r - 1);
    } else {

    }*/
    printf("%lld\n", l + r - 1);
    return 0;
}