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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <array>
#include <algorithm>
#include <cstring>
#include <queue>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <cstdlib>
#include <sstream>
#include <bitset>

using namespace std;

using ll = long long;
using Vec = vector<ll>;
using P = pair<ll, ll>;
using VecP = vector<P>;
#define rep(i, n) for(ll i=0;i<(n);i++)
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))

//static const ll MOD = 1000000007;
//static const ll INF = 1000000000000000000;
//#define PI 3.14159265359

int main() {
    // input
    ll N;
    cin >> N;
    Vec A(N), B(N);
    rep(i, N) {
        cin >> A[i] >> B[i];
    }

    //process
    sort(A.begin(), A.end());
    sort(B.begin(), B.end());
    ll ans = 0;
    if (N % 2 == 0) {
        ll center = N / 2 - 1;
        double center_A = (double) (A[center] + A[center + 1]) / 2;
        double center_B = (double) (B[center] + B[center + 1]) / 2;
        ans = (center_B - center_A + 0.5) * 2;
    } else {
        ll center = N / 2;
        ans = B[center] - A[center] + 1;
    }

    cout << ans << endl;

    return 0;
}