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>
# define rr
# define int long long
const int N=200010,INF=0x3f3f3f3f;
int A[N],B[N];
int n;
inline int read(void){
	int res,f=1;
	char c;
	while((c=getchar())<'0'||c>'9')
		if(c=='-')f=-1;
	res=c-48;
	while((c=getchar())>='0'&&c<='9')
		res=res*10+c-48;
	return res*f;
}
signed main(void){
	n=read();
	for(rr int i=1;i<=n;++i){
		A[i]=read(),B[i]=read();
	}
	std::sort(A+1,A+1+n);
	std::sort(B+1,B+1+n);
/*	for(rr int i=1;i<=n;++i){
		printf("%lld ",A[i]);
	}
	puts("");
	for(rr int i=1;i<=n;++i){
		printf("%lld ",B[i]);
	}*/
	if(n%2){
		printf("%lld",B[n/2+1]-A[n/2+1]+1);
		return 0;
	}else{
		int ret=(B[n/2+1]-A[n/2+1]+1)*(B[n/2]-A[n/2]+1);
		if(A[n/2+1]<=B[n/2]){
			ret-=(B[n/2]-A[n/2+1]+1);
		}
		printf("%lld",ret);
		return 0;
	}
	return 0;
}