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>

#define LL long long int

typedef struct {
    int index;
    LL value;
} tValueLL;

int cmpLL(const void *n1, const void *n2) {

    if (((tValueLL *)n1)->value > ((tValueLL *)n2)->value) {
        return 1;
    } else if (((tValueLL *)n1)->value < ((tValueLL *)n2)->value) {
        return -1;
    }

    return 0;
}


int main(void) {

    int n;
    scanf("%d", &n);
    tValueLL *a = (tValueLL *)malloc(n * sizeof(tValueLL));
    tValueLL *b = (tValueLL *)malloc(n * sizeof(tValueLL));
    for (int i = 0; i < n; i ++) {
        a[i].index = i;
        b[i].index = i;
        scanf("%lld %lld", &a[i].value, &b[i].value);
    }

    qsort(a, n, sizeof(tValueLL), cmpLL);
    qsort(b, n, sizeof(tValueLL), cmpLL);

    int ret;
    if (n % 2 == 1) {
        ret = b[(n - 1) / 2].value - a[(n - 1) / 2].value + 1;
    } else {
        ret = (b[n / 2 - 1].value + b[n / 2].value) - (a[n / 2 - 1].value + a[n / 2].value) + 1;
    }

    printf("%d", ret);

    return 0;
}