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;
using int64 = long long;
#define int int64
#define rep(i, n) for(int i=0; i<n; i++)
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define SORT(x) sort(x.begin(), x.end())
#define GSORT(x) sort(x.begin(), x.end(), greater<int>())
#define mk make_pair
#define fi first
#define se second
#define pb push_back
#define ALL(x) x.begin(), x.end()
#define V(T) vector<T>
typedef pair<int, int> P;
typedef pair<P, P> PP;
typedef vector<int> vi;
typedef vector<vi> vvi;
int max(int a, int b) {if(b>a) return b; else return a;}
int min(int a, int b) {if(b<a) return b; else return a;}
const int INF = 1e18;

signed main() {
    int N;
    cin >> N;
    vi A(N), B(N);
    rep(i, N) {
        cin >> A[i] >> B[i];
    }
    SORT(A); SORT(B);

    if(N%2) cout << B[N/2]-A[N/2]+1 << endl;
    else {
        double at = double(A[N/2-1]+A[N/2])/2.0;
        double bt = double(B[N/2-1]+B[N/2])/2.0;
        cout << (bt-at)*2+1 << endl;
    }
    return 0;
}