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;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> PL;
#define int ll

const ll MOD = 1000000007;
const ll INF_LL = (ll)1000000007 * 1000000007;
const int INF_INT = (int)1000000007;
const double PI = 3.14159265358979323846;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define irep(itr, st) for (auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for (auto itr = (st).rbegin(); itr != (st).rend(); ++itr)

#define m0(x) memset((x), 0, sizeof((x)))
#define m1(x) memset((x), -1, sizeof((x)))

// xにはvectorなどのコンテナ
#define ALL(x) (x).begin(), (x).end()  // sortなどの引数を省略したい
#define RALL(x) (x).rbegin(), (x).rend()  // sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size())     // sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x))  //最大値を求める
#define MIN(x) *min_element(ALL(x))  //最小値を求める
#define all(x) (x).begin(), (x).end()

//略記
#define pb emplace_back  // vectorヘの挿入
#define mp make_pair     // pairのコンストラクタ
#define F first          // pairの一つ目の要素
#define S second         // pairの二つ目の要素

#define BITCOUNT __builtin_popcount
#define BITCOUNT_LL(x) __builtin_popcountll(x)

#define perm(c)             \
  sort(all(c));             \
  for (bool c##p = 1; c##p; \
       c##p = next_permutation(all(c)))  //順列 123 132 213 231 312 321

#define BIT(n) (1LL << (n))

#ifdef DEBUG
#define PRINT(A) std::cout << (#A) << ": " << (A) << std::endl;
#else
#define PRINT(A)
#endif

//入力高速化
struct IoSetup {
  IoSetup() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;

template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

signed main() {
  int N; cin >> N;
  vector<int> A(N), B(N);
  rep(i, N) {
    cin >> A[i];
    cin >> B[i];
  }
  sort(all(A));
  sort(all(B));

  if(N%2==1){
    cout << B[N/2]-A[N/2]+1 << endl;
  }
  else{
    cout << (B[N/2]+B[N/2-1]) - (A[N/2]+A[N/2-1]) + 1 << endl;
  }
}