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"
#define int long long
#define float double
#define double long double
#define endl "\n"
#define debug cout<<"Good uptill here"<<endl<<flush;
#define deb(x) cout<<#x<<":"<<x<<endl;

using namespace std;

int32_t main(){
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	//CODE FOR SPECIFIED INPUT AND OUTPUT
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	int n;
	cin>>n;
	int mini[n], maxi[n];
	for(int i=0;i<n;i++){
		cin>>mini[i]>>maxi[i];
	}
	sort(mini,mini+n);
	sort(maxi,maxi+n);
	double l,h;
	int ans;
	if(n%2==0){
		l = (double)(mini[n/2] + mini[n/2-1])/2;
		h = (double)(maxi[n/2] + maxi[n/2-1])/2;
		ans = (double)((h-l)*2 +1);
	}else{
		l = mini[n/2];
		h = maxi[n/2];
		ans = (h-l)+1;
	}
	cout<<ans<<endl;
	return 0;
}