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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>
#define N 200001
#include <stdlib.h>

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


int main()//リニューアル
{

    int a;

    long long Big[N];
    long long small[N];

    scanf("%d", &a);

    for (int i = 0; i < a; i++)
    { //道を読み取る

        scanf("%lld %lld", &small[i],&Big[i]);

    }
    qsort(Big, a, sizeof(long long), compare);
    qsort(small,a, sizeof(long long), compare);

    if(a%2==0){
        long long distance =(Big[a / 2] + Big[a / 2-1] )- (small[a / 2]+small[a / 2-1]);//わる２して普通に差
        printf("%lld", distance+1);
    }
    else{//基
        printf("%lld", Big[a / 2] - small[a / 2] + 1);
    }


    printf("\n");
}