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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less_equal <int>, rb_tree_tag, tree_order_statistics_node_update>ordered_set;
using ll = long long;
using ld = long double;
#define all(x) x.begin(), x.end()
#define ff first
#define ss second
#define rep(i,n) for (ll i = 0; i < n; ++i)
#define forn(i,a,b) for (ll i =a; i <=b; ++i)
void debug()
{
    cerr << "\n";
}
template<typename Head, typename... Tail>
void debug(Head H, Tail... T)
{
    cerr << H << " ";
    debug(T...);
}
#define debug(...) cerr<<"["<<#__VA_ARGS__<<"] : ";debug(__VA_ARGS__);
const ll maxn = 1e6 + 10;
void solve()
{
    ll n;
    cin>>n;
    ll a[n],b[n];
    rep(i,n)
    cin>>a[i]>>b[i];
    sort(a,a+n);
    sort(b,b+n);
    if(n&1)
    {
        cout<<b[n/2]-a[n/2]+1<<"\n";
    }
    else
    {
        cout<<(b[n/2]+b[n/2-1]-a[n/2]-a[n/2-1]+1)<<"\n";
    }
}
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ll t=1;
    // cin >> t;
    rep(i, t)
    {
        solve();
    }
    return 0;
}