codeforces364D

Ghd

 CodeForces - 364D 

John Doe offered his sister Jane Doe find the gcd of some set of numbers a.

Gcd is a positive integer g, such that all number from the set are evenly divisible by g and there isn't such g' (g' > g), that all numbers of the set are evenly divisible by g'.

Unfortunately Jane couldn't cope with the task and John offered her to find the ghd of the same subset of numbers.

Ghd is a positive integer g, such that at least half of numbers from the set are evenly divisible by g and there isn't such g' (g' > g) that at least half of the numbers from the set are evenly divisible by g'.

Jane coped with the task for two hours. Please try it, too.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁶) showing how many numbers are in set a. The second line contains space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10¹²). Please note, that given set can contain equal numbers.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the %I64d specifier.

Output

Print a single integer g — the Ghd of set a.

Examples

Input

6
6 2 3 4 5 6

Output

3

Input

5
5 5 6 10 15

Output

5

题意:n个数中取一半，使得gcd最大

sol：似乎是鬼畜的随机化算法，每次随机取一个，然后计算所有数字与它的gcd，再随便判断一下个数是否满足一半就好了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0; bool f=0; char ch=' ';
    while(!isdigit(ch))    {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<3)+(s<<1)+(ch^48); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0) {putchar('-'); x=-x;}
    if(x<10) {putchar(x+'0'); return;}
    write(x/10); putchar((x%10)+'0');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=1000005;
int n,m;
ll a[N],cnt=0,b[N],ans=0;
struct Node{ll num; int cnt;}c[N];
inline ll gcd(ll a,ll b)
{
    return (!b)?(a):(gcd(b,a%b));
}
inline void Solve()
{
    ll tmp=a[rand()%n+1];
//    cout<<"tmp="<<tmp<<endl;
    int i,j;
    for(i=1;i<=n;i++) b[i]=gcd(a[i],tmp);
    sort(b+1,b+n+1);
    cnt=0; c[++cnt].num=b[1]; c[cnt].cnt=1;
    for(i=2;i<=n;i++)
    {
        if(b[i]!=b[i-1]){c[++cnt].num=b[i]; c[cnt].cnt=0;} c[cnt].cnt++;
    }
    for(i=1;i<=cnt;i++)
    {
        int sum=0;
        for(j=1;j<=cnt;j++) if(c[j].num%c[i].num==0) sum+=c[j].cnt;
        if(sum>=m) ans=max(ans,c[i].num);
    }
}
int main()
{
    srand(20030310);
    int i;
    R(n);
    for(i=1;i<=n;i++) R(a[i]);
    m=(n+1)/2;
    for(i=1;i<=10;i++) Solve();
    Wl(ans);
    return 0;
}
/*
input
6
6 2 3 4 5 6
output
3

input
5
5 5 6 10 15
output
5
*/