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>

#define int long long
#define ull unsigned long long
#define ld long double
#define rep(a) rep1(i,a)
#define rep1(i,a) rep2(i,0,a)
#define rep2(i,b,a) for(int i=(b); i<((int)(a)); i++)
#define rep3(i,b,a) for(int i=(b); i>=((int)(a)); i--)
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define sz(a) ((int)a.size())
#define pow2(x) (1ll<<(x))
#define ceiling(a,b) (((a)+(b)-1)/(b))
#define print0(a) cout << (a) << ' '
#define print1(a) cout << (a) << '\n'
#define print2(a,b) cout << (a) << ' ',print1(b)
#define print3(a,b,c) cout << (a) << ' ',print2(b,c)
#define print4(a,b,c,d) cout << (a) << ' ',print3(b,c,d)
#ifdef i_am_noob
#define debug(a) print2(#a,a)
#else
#define debug(a)
#endif
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>

using namespace std;
using namespace __gnu_pbds;

const int Mod=1000000007,Mod2=998244353;
const int MOD=Mod;

inline int add(int a, int b){return (a+b>=MOD?a+b-MOD:a+b);}
inline int sub(int a, int b){return (a+MOD-b>=MOD?a-b:a+MOD-b);}
int Pow(int a, int b){
	if(!b) return 1;
	int tmp=Pow(a*a%MOD,b>>1);
	return (b&1?tmp*a%MOD:tmp);
}
inline int inverse(int n){return Pow(n,MOD-2);}

const int maxn=200005;
//i_am_noob
int n,a[maxn],b[maxn];

signed main(){
	ios_base::sync_with_stdio(0),cin.tie(0);
	cin >> n;
	rep(n) cin >> a[i] >> b[i];
	sort(a,a+n),sort(b,b+n);
	if(n%2==1) print1(b[n>>1]-a[n>>1]+1);
	else print1(b[n/2]+b[n/2-1]-a[n/2]-a[n/2-1]+1);
	return 0;
}