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)

//set many funcs template
//Ver.20190820
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#include<assert.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
int ceil(int a,int b){if(a%b==0){return a/b;}return (a/b)+1;}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int nHr(int a,int b){return nCr(a+b-1,b);}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}
int sankaku(int x){return ((1+x)*x)/2;}
void swap(int *a,int *b){int c;c=(*a);(*a)=(*b);(*b)=c;}
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}
void llswap(long long *a,long long *b){long long c;c=(*a);(*a)=(*b);(*b)=c;}
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}
void dbswap(double *a,double *b){double c;c=(*a);(*a)=(*b);(*b)=c;}
void chswap(char *a,char *b){char c;c=(*a);(*a)=(*b);(*b)=c;}
int sortfncsj(const void *a,const void *b){if(*(int *)a>*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int sortfnckj(const void *a,const void *b){if(*(int *)a<*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int llsortfncsj(const void *a,const void *b){if(*(long long *)a>*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int llsortfnckj(const void *a,const void *b){if(*(long long *)a<*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int dbsortfncsj(const void *a,const void *b){if(*(double *)a>*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int dbsortfnckj(const void *a,const void *b){if(*(double *)a<*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int strsortfncsj(const void *a,const void *b){return strcmp((char *)a,(char *)b);}
int strsortfnckj(const void *a,const void *b){return strcmp((char *)b,(char *)a);}
int chsortfncsj(const void *a,const void *b){if(*(char *)a>*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}
int chsortfnckj(const void *a,const void *b){if(*(char *)a<*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}

void shuffledget(int x[],int n){
    int i,b[524288],p,c;
    for(i=0;i<n;i++){
        b[i]=i;
    }
    for(i=n;i>=1;i--){
        p=rand()%i;
        c=b[i-1];b[i-1]=b[p];b[p]=c;
    }
    for(i=0;i<n;i++){
        scanf("%d",&x[b[i]]);
    }
}

int dx4[4]={1,-1,0,0};
int dy4[4]={0,0,1,-1};
int dx8[8]={-1,-1,-1,0,0,1,1,1};
int dy8[8]={-1,0,1,-1,1,-1,0,1};

int search(int x,int a[],int n){
    int st=0,fi=n-1,te;
    while(st<=fi){
        te=(st+fi)/2;
        if(a[te]<x){st=te+1;}else{fi=te-1;}
    }
    return st;
}

void prarr(int arr[],int n){
  int i;
  for(i=0;i<n;i++){
    if(i){printf(" ");}
    printf("%d",arr[i]);
  }
  printf("\n");
  return;
}

void getperm(int a[],int n){
  int i,p;
  for(i=0;i<n;i++){
    a[i]=i;
  }
  for(i=n-1;i>=1;i--){
    p=rand()%(i+1);
    swap(&a[p],&a[i]);
  }
}

long long a[524288],b[524288];

int main(void){
  long long i,j,n,m,k,c,h,w,r=0,l,t;
  long long st,fi;
  scanf("%lld",&n);
  for(i=0;i<n;i++){
    scanf("%lld%lld",&a[i],&b[i]);
  }
  qsort(a,n,sizeof(long long),llsortfncsj);
  qsort(b,n,sizeof(long long),llsortfncsj);
  if(n%2==1){
    st=a[n/2];
    fi=b[n/2];
    printf("%lld\n",fi-st+1);
  }
  else{
    st=a[(n/2)-1]+a[n/2];
    fi=b[(n/2)-1]+b[n/2];
    printf("%lld\n",fi-st+1);
  }
  return 0;
}