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 ll long long
#define loop(__x, __start, __end) for(int __x = __start; __x < __end; __x++)


int main() {
  int n; cin >> n;
  vector<ll> A(n), B(n);
  loop(i,0,n) cin >> A[i] >> B[i];
  sort(A.begin(), A.end());
  sort(B.begin(), B.end());
  if (n&1) {
    ll x = A[n/2];
    ll y = B[n/2];
    cout << y-x+1 << endl;
    return 0;
  }

  ll x2 = A[(n/2)-1] + A[n/2];
  ll y2 = B[(n/2)-1] + B[n/2];
  ll ans = (y2/2 - (x2/2 + (x2%2!=0))) * 2 + 1;
  ans += x2&1;
  ans += y2&1;
  cout << ans << endl;
  return 0;
}