Problem:
Score : 700 points

Problem Statement

There are N bags, each containing two white balls. The i-th box contains two balls with integers x_i and y_i written on them, respectively.

For each of these bags, you will paint one of the balls red, and paint the other blue.

Afterwards, the 2N balls will be classified according to color.

Then, we will define the following:

R_{max}: the maximum integer written on a ball painted in red

R_{min}: the minimum integer written on a ball painted in red

B_{max}: the maximum integer written on a ball painted in blue

B_{min}: the minimum integer written on a ball painted in blue

Find the minimum possible value of (R_{max} - R_{min}) \times (B_{max} - B_{min}).

Constraints

1 ≤ N ≤ 200,000

1 ≤ x_i, y_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the minimum possible value.

Sample Input 1

3
1 2
3 4
5 6

Sample Output 1

15

The optimal solution is to paint the balls with x_1, x_2, y_3 red, and paint the balls with y_1, y_2, x_3 blue.

Sample Input 2

3
1010 10
1000 1
20 1020

Sample Output 2

380

Sample Input 3

2
1 1
1000000000 1000000000

Sample Output 3

999999998000000001

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <fstream>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <queue>
#include <random>

using namespace std;

#define min(a,b) ((a)<(b) ? (a):(b))
#define max(a,b) ((a)>(b) ? (a):(b))

long long maxl(long long a,long long b){
    if(a>=b){
        return a;
    }
    return b;
}
long long minl(long long a,long long b){
    if(a<=b){
        return a;
    }
    return b;
}

long long gcd(long long a, long long b){
    if(a<b){
        swap(a,b);
    }
    while(b){
        long long r = a%b;
        a=b;
        b=r;
    }
    return a;
}

long long lcm(long long a, long long b){
    return (a*b)/gcd(a,b);
}

int isPrim(int a){
    if(a==1){
        return 0;
    }
    for(int i=2;i<=(a+1)/2;i++){
        if(a%i==0){
            return 0;
        }
    }
    return 1;
}

long long mod_pow(long long x, long long n, long long mod){
    //xのn乗を計算するのにn乗を2進表記にして計算
    //x^22 = x^16 + x^4 + x^2
    long long ret=1;
    while(n>0){
        if(n%2==1){
            ret=(ret*x)%mod;//答えに付加
        }
        x=(x*x)%mod;//2乗
        n=n/2;
    }
    return ret;
}


struct XX{
    int u;
    int v;
    int i;
};

class xxIntu {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        //第2条件
        if((riLeft.u) == (riRight.u)){
            return riLeft.i < riRight.i;//<:昇順(小さいものから順番)、>:降順(大きいものから順番)
        }
        //第1条件
        return (riLeft.u) > (riRight.u);
    }
};
//bool operator< (const XX &riLeft, const XX &riRight){
//    return riLeft.d < riRight.d;
//};



int ppar[200001];
int rrank[200001];

void init(int n){
    for(int i=1;i<=n;i++){
        ppar[i]=i;
        rrank[i]=0;
    }
}

int find(int x){
    if(ppar[x]==x){
        return x;
    }else{
        return ppar[x]=find(ppar[x]);
    }
}

void unite(int x,int y){
    x=find(x);
    y=find(y);
    if(x==y){
        return;
    }
    if(rrank[x]<rrank[y]){
        ppar[x]=ppar[y];
    }else{
        ppar[y]=x;
        if(rrank[x]==rrank[y]){
            rrank[x]++;
        }
    }
}
bool same(int x,int y){
    return find(x)==find(y);
}

//kruskal
struct edge{
    int u;
    int v;
    int cost;
};

bool comp(edge& e1,edge& e2){
    return e1.cost < e2.cost;
}

edge es[2000];
int V,E;
int ans=0;
int kruskal(){
    sort(es,es+E,comp);
    init(V);
    int res = 0;
    for(int i=0;i<E;i++){
        edge e = es[i];
        if(!same(e.u,e.v)){
            unite(e.u,e.v);
            res+=e.cost;
            ans++;
        }
    }
    return res;
}


int main(int argc, const char * argv[])
{
    //std::ios::sync_with_stdio(false);
    //scanf("%s",S);
    //scanf("%d",&N);
    //sscanf(tmp.c_str(),"%dd%d%d",&time[i], &dice[i], &z[i]);
    //getline(cin, target);
    //cin >> x >> y;
    //テスト用
    //ifstream ifs( "1_06.txt" );
    //ifs >> a;
    //ここから
    
    int N;
    cin >> N;
    int x[200000];
    int y[200000];
    for(int i=0;i<N;i++){
        cin >> x[i] >> y[i];
    }
    for(int i=0;i<N;i++){
        if(x[i]<y[i]){
            int tmp=x[i];
            x[i]=y[i];
            y[i]=tmp;
        }
    }
    long long maxx=0;
    long long minx=2000000000;
    for(int i=0;i<N;i++){
        if(maxx<x[i]){
            maxx=x[i];
        }
        if(minx>x[i]){
            minx=x[i];
        }
    }
    long long maxy=0;
    long long miny=2000000000;
    for(int i=0;i<N;i++){
        if(maxy<y[i]){
            maxy=y[i];
        }
        if(miny>y[i]){
            miny=y[i];
        }
    }
    long long ans=(maxx-minx)*(maxy-miny);
    if((maxx-miny)*(maxy-minx)<ans && maxx-miny>=0 && maxy-minx>0){
        ans=(maxx-miny)*(maxy-minx);
    }
    cout << ans << endl;

    //ここまで
    //cout << "ans" << endl;改行含む
    //printf("%.0f\n",ans);//小数点以下表示なし
    //printf("%.7f\n",p);
    //printf("%f\n",pow(2,ans.size()));
    
    return 0;
}