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,n) for(int i=0; i<(n); i++)
#define FOR(i,a,n) for(int i=(a); i<(n); i++)
#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define sz(x) int(x.size())
using namespace std;
using ll = long long;
using P = pair<int ,int>;
const int INF = 1001001001;
const int MOD = 1000000007;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

#ifdef _DEBUG
#include "debug.h"
#else
#define debug(...)
#define print(...)
#endif

int main() {
    int n; cin >> n;
    vector<P> a(n);
    rep(i, n) {
        int x, y; cin >> x >> y;
        a[i] = {x, y};
    }
    sort(all(a));
    int left;
    if (n>2) left = a[1].fi;
    else left = a[0].fi;
    sort(all(a), [](P &a, P& b) {return a.se<b.se;});
    int right;
    if (n>2) right = a[n-2].se;
    else right = a[n-1].se;
    debug(left, right);
    int ans;
    if (n%2) ans = right - left + 1;
    else if (n==2) {
        int left2 = a[1].fi, right2 = a[n-2].se;
        int left3 = left2+left, right3 = right2+right;
        ans = right3 - left3 + 1;
    }
    else ans = (right-left+1)+(right-left);
    cout << ans << endl;
}