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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { std::cout << #a << " = " << a << "\n"; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }
template<class T, class V> istream& operator>> (istream& ist, pair<T, V>& p) { return ist >> p.first >> p.second; }
template<class T> ostream& operator<< (ostream& ost, pair<T, T>& p) { return ost << p.first << ", " << p.second; }
template<class T> istream& operator>> (istream& ist, vector<T>& vs) { for(auto& e: vs) ist >> e; return ist; }

typedef long long ll;
int const inf = INT_MAX / 2;

int main() {
  int N; cin >> N;
  vector<int> A, B;
  rep(_, N) {
    int a, b; cin >> a >> b;
    A.push_back(a);
    B.push_back(b);
  }
  sort(A.begin(), A.end());
  sort(B.begin(), B.end());
  if (N % 2) cout << (B[N/2] - A[N/2] + 1) << "\n";
  else cout << (B[N/2-1] + B[N/2] - A[N/2-1] - A[N/2] + 1) << "\n";
}