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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long

int n;
int a[200100];
int b[200100];

int compare_int(const void *a, const void *b){
    return *(int*)a - *(int*)b;
}

int main(){
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    scanf("%d%d",&a[i],&b[i]);}
  qsort(a,n,sizeof(int),compare_int);
  qsort(b,n,sizeof(int),compare_int);
  if(n%2==0){
    printf("%d\n",b[n/2]+b[n/2-1]-a[n/2]-a[n/2-1]+1);}
  else{
    printf("%d\n",b[n/2]-a[n/2]+1);}
  
  
  return 0;
}