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>
using namespace std;

#define pb push_back
#define eb emplace_back
#define mk make_pair
#define fi first
#define se second
#define mset(a, b) memset(a, b, sizeof(a))
#define DBG(x) cout << "[" << #x << "]: " << x << endl
using ll = long long;
using pii = pair<int, int>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<class Ty> Ty randint(Ty a, Ty b) { return uniform_int_distribution<Ty>(a, b)(rng); }
template<class num> inline void rd(num& x) {
	char c; while(isspace(c = getchar()));
	bool neg = false;
	if(!isdigit(c)) neg = (c == '-'), x = 0;
	else x = c - '0';
	while(isdigit(c = getchar())) x = (x << 3) + (x << 1) + c - '0';
	x = neg ? -x : x; }
template <class... Args> inline void rd(Args&... args) { (rd(args), ...); }
template <class... Args> inline void rf(Args&... args) { (cin >> ... >> args); }
template <class... Args> inline void pr(Args&&... args) { (cout << ... << args); }

const int MAXN = 2e5 + 5, INF = 0x3f3f3f3f;

ll a[MAXN];
ll b[MAXN];

int main() {
	int n; rd(n);
	for (int i = 1; i <= n; i++)
		rd(a[i], b[i]);
	sort(a+1, a+n+1);
	sort(b+1, b+n+1);

	ll ans = 0;
	if (n%2 == 0) {
		ll lo = a[n/2] + a[n/2+1];
		ll hi = b[n/2] + b[n/2+1];

		ans = hi/2 - lo/2;
		ans += lo%2;
		ans += hi%2;
	} else {
		ans = b[n/2+1] - a[n/2+1] + 1;
	}

	printf("%lld\n", ans);

}