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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define rep(i,k,n) for(int (i)=(k);(i)<(n);(i)++)
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
#define abs(a) (a)>0 ?(a):-(a)
#define llong long long

int main(){
	int N; scanf("%d",&N);
	llong x[N+1];
	llong up[N+2],down[N+2];
	rep(i,0,N+1){
		x[i]=0;
		up[i]=0;
		down[i]=0;
	}
	up[N+2]=0; down[N+2]=0;
	rep(i,0,N){
		int a,b; scanf("%d %d",&a,&b);
		rep(j,a,b+1) x[j]++;
	}
	rep(i,0,N){
		rep(j,0,N){
			up[N-i] = up[N+1-i]+x[N-i];
			down[i+1] = down[i]+x[i+1];
		}
	}
	llong M = down[N];
	llong cnt=0;
	rep(i,0,N){
		if( down[i]*2 <= M && up[i]*2 >= M ) cnt++;
		else if( down[i]*2 >= M && up[i]*2 <= M ) cnt++;
		if( down[i]*2+1==M || down[i]*2-1==M ) cnt++;
		if( up[i]*2+1==M || up[i]*2-1==M ) cnt++;
	}
	printf("%lld",cnt);
}