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: Wrong Answer)

//#include <bits/stdc++.h>
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <map>
#include <iomanip>
#include <stdlib.h>
#include <stdio.h>
#include <queue>
#include <deque>
#include <set>
#include <stack>
#include <time.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;

#define fi first
#define se second
#define mp make_pair
 
const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const ll MOD3 = 1812447359;
const ll INF = 1ll << 62;
const double PI = 2 * asin(1);

void yes() {printf("yes\n");}
void no() {printf("no\n");}
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

int N;
int A[int(2e5 + 5)], B[int(2e5 + 5)];


int main(){
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d%d", A + i, B + i);

  int minL = 0, minR = 1e9 + 1;
  while (minR - minL > 1){
    int now = (minL + minR) / 2, cnt = 0;
    for (int i = 0; i < N; i++) {
      if (B[i] < now) cnt++;
    }
    if (cnt <= (N - 1) / 2) minL = now;
    else minR = now;
  }

  int minL2 = 0, minR2 = 1e9 + 1;
  while (minR2 - minL2 > 1){
    int now = (minL2 + minR2) / 2, cnt = 0;
    for (int i = 0; i < N; i++){
      if (now < A[i]) cnt++;
    }
    if (cnt <= (N - 1) / 2) minR2 = now;
    else minL2 = now;
  }

  bool flag = true;
  for (int i = 0; i < N - 1; i++){
    if (A[i] != A[i + 1] && B[i] != B[i + 1]) flag = false;
  }


  if (N % 2 != 0){
    cout << minL - minL2 << endl;
  }else{
    if (flag) cout << (minL - minL2) * 2 - 1 << endl;
    else cout << (minL - minL2) * 2 + 1 << endl;
  }

  return 0;
}