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)

// https://atcoder.jp/contests/abc169/tasks/abc169_e

#include "algorithm"
#include "bitset"
#include "cmath"
#include "functional"
#include "iomanip"
#include "iostream"
#include "map"
#include "numeric"
#include "queue"
#include "set"
#include "string"
#include "vector"
#define rep(i, to) for (ll i = 0; i < (to); ++i)
#define rep1(i, to) for (ll i = 1; i <= (to); ++i)
#define repf(i, from, to) for (ll i = from; i < (to); ++i)
#define repr(i, from) for (ll i = from - 1; i >= 0; --i)
#define all(vec) vec.begin(), vec.end()
#define unless(cond) if (!(cond))
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef long double ld;
template <typename T>
using V = vector<T>;
using VL = V<ll>;
using VVL = V<VL>;
template <typename T, typename U>
using P = pair<T, U>;
using PL = P<ll, ll>;
using VPL = V<PL>;

template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
inline ll len(V<T> arr) {
  return arr.size();
}

struct exit_exception : public std::exception {
  const char* what() const throw() { return "Exited"; }
};

template <typename T>
void drop(T res) {
  cout << res << endl;
  throw exit_exception();
}

const ll INF = 1e18;

void solve();

#ifndef TEST
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  try {
    solve();
  } catch (exit_exception& e) {
  }

  return 0;
}
#endif

void solve() {
  ll N;
  cin >> N;

  VL ls(N), rs(N);
  rep(i, N) cin >> ls[i] >> rs[i];

  ll res;

  if (N % 2 == 1) {
    ll mi = ls[N / 2];
    ll ma = rs[N / 2];
    res = ma - mi + 1;
  } else {
    ll mi_doubled = ls[N / 2 - 1] + ls[N / 2];
    ll ma_doubled = rs[N / 2 - 1] + rs[N / 2];
    res = ma_doubled - mi_doubled + 1;
  }

  cout << res << "\n";
}