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 MAXLEN 2*10000


int mycmp(const void* aa, const void* bb) {
	const unsigned int* a = aa;
	const unsigned int* b = bb;
	if (*a > *b) {
		return 1;
	} else if (*a < *b) {
		return -1;
	} else {
		return 0;
	}
}

int main(void) {
	size_t n;
	scanf("%zu", &n);

	/* 以下はstack overflow を起こすのでmallocで動的に確保しよう*/
	/*
	unsigned int as[MAXLEN], bs[MAXLEN];
	for (size_t i = 0; i < MAXLEN; i++) {
		as[i] = -1;
		bs[i] = -1;
	}
	*/

	unsigned int* as = malloc(sizeof(unsigned int) * n);
	unsigned int* bs = malloc(sizeof(unsigned int) * n);

	for (size_t i = 0; i < n; i++) {
		scanf("%u %u", as+i, bs+i);
	}

	qsort(as, n, sizeof(unsigned int), mycmp);
	qsort(bs, n, sizeof(unsigned int), mycmp);

	unsigned int ans;
	
	if (n % 2 == 1) {
		unsigned int a, b;
		a = as[(n-1)/2];
		b = bs[(n-1)/2];
		ans = b - a + 1;
	} else {
		unsigned int a0, a1, b0, b1;
		a0 = as[(n-1)/2];
		a1 = as[n/2];
		b0 = bs[(n-1)/2];
		b1 = bs[n/2];
		ans = b0 - a0 + b1 - a1 + 1;
	}
	
	printf("%u\n", ans);
	//fprintf(stderr, "OK!\n");
	return 0;
}