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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define ss second
#define ff first
#define N 100005
#define inf 1000000009
#define ll long long
#define pii pair<int,int>
#define sz(a) int(a.size())
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

int n,a[N],b[N];

int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++)
		scanf("%d%d",&a[i],&b[i]);		
	sort(a+1,a+n+1);
	sort(b+1,b+n+1);
	if(n % 2 == 1) printf("%d",b[(n+1)/2]-a[(n+1)/2]+1);
	else {
		int ma = a[n/2+1] + a[n/2];
		int mb = b[n/2+1]+b[n/2];
		printf("%d",mb - ma + 1);
	}
}