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)

#pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
void *wmem;
char memarr[96000000];
template<class T> inline void walloc1d(T **arr, int x, void **mem = &wmem){
  static int skip[16] = {0, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
  (*mem) = (void*)( ((char*)(*mem)) + skip[((unsigned long long)(*mem)) & 15] );
  (*arr)=(T*)(*mem);
  (*mem)=((*arr)+x);
}
template<class T1> void sortA_L(int N, T1 a[], void *mem = wmem){
  sort(a, a+N);
}
inline int my_getchar_unlocked(){
  static char buf[1048576];
  static int s = 1048576;
  static int e = 1048576;
  if(s == e && e == 1048576){
    e = fread_unlocked(buf, 1, 1048576, stdin);
    s = 0;
  }
  if(s == e){
    return EOF;
  }
  return buf[s++];
}
inline void rd(int &x){
  int k;
  int m=0;
  x=0;
  for(;;){
    k = my_getchar_unlocked();
    if(k=='-'){
      m=1;
      break;
    }
    if('0'<=k&&k<='9'){
      x=k-'0';
      break;
    }
  }
  for(;;){
    k = my_getchar_unlocked();
    if(k<'0'||k>'9'){
      break;
    }
    x=x*10+k-'0';
  }
  if(m){
    x=-x;
  }
}
inline void rd(long long &x){
  int k;
  int m=0;
  x=0;
  for(;;){
    k = my_getchar_unlocked();
    if(k=='-'){
      m=1;
      break;
    }
    if('0'<=k&&k<='9'){
      x=k-'0';
      break;
    }
  }
  for(;;){
    k = my_getchar_unlocked();
    if(k<'0'||k>'9'){
      break;
    }
    x=x*10+k-'0';
  }
  if(m){
    x=-x;
  }
}
struct MY_WRITER{
  char buf[1048576];
  int s;
  int e;
  MY_WRITER(){
    s = 0;
    e = 1048576;
  }
  ~MY_WRITER(){
    if(s){
      fwrite_unlocked(buf, 1, s, stdout);
    }
  }
}
;
MY_WRITER MY_WRITER_VAR;
void my_putchar_unlocked(int a){
  if(MY_WRITER_VAR.s == MY_WRITER_VAR.e){
    fwrite_unlocked(MY_WRITER_VAR.buf, 1, MY_WRITER_VAR.s, stdout);
    MY_WRITER_VAR.s = 0;
  }
  MY_WRITER_VAR.buf[MY_WRITER_VAR.s++] = a;
}
inline void wt_L(char a){
  my_putchar_unlocked(a);
}
inline void wt_L(long long x){
  int s=0;
  int m=0;
  char f[20];
  if(x<0){
    m=1;
    x=-x;
  }
  while(x){
    f[s++]=x%10;
    x/=10;
  }
  if(!s){
    f[s++]=0;
  }
  if(m){
    my_putchar_unlocked('-');
  }
  while(s--){
    my_putchar_unlocked(f[s]+'0');
  }
}
int N;
long long A[200000];
long long B[200000];
int main(){
  wmem = memarr;
  rd(N);
  {
    int Lj4PdHRW;
    for(Lj4PdHRW=(0);Lj4PdHRW<(N);Lj4PdHRW++){
      rd(A[Lj4PdHRW]);
      rd(B[Lj4PdHRW]);
    }
  }
  sortA_L(N,A);
  sortA_L(N,B);
  if(N%2==1){
    wt_L(B[N/2]-A[N/2]+ 1);
    wt_L('\n');
  }
  else{
    wt_L(B[N/2]+B[(N-1)/2]-A[N/2]-A[(N-1)/2]+ 1);
    wt_L('\n');
  }
  return 0;
}
// cLay varsion 20200509-1

// --- original code ---
// int N;
// ll A[2d5], B[2d5];
// {
//   rd(N,(A,B)(N));
//   sortA(N,A);
//   sortA(N,B);
//   wt(if[N%2==1, B[N/2]-A[N/2], B[N/2]+B[(N-1)/2]-A[N/2]-A[(N-1)/2]] + 1);
// }