A. Orac and LCM

For the multiset of positive integers s={s1,s2,…,sk}s={s1,s2,…,sk}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of ss as follow:

• gcd(s)gcd(s) is the maximum positive integer xx, such that all integers in ss are divisible on xx.
• lcm(s)lcm(s) is the minimum positive integer xx, that divisible on all integers from ss.

For example, gcd({8,12})=4,gcd({12,18,6})=6gcd({8,12})=4,gcd({12,18,6})=6 and lcm({4,6})=12lcm({4,6})=12. Note that for any positive integer xx, gcd({x})=lcm({x})=xgcd({x})=lcm({x})=x.

Orac has a sequence aa with length nn. He come up with the multiset t={lcm({ai,aj}) | i<j}t={lcm({ai,aj}) | i<j}, and asked you to find the value of gcd(t)gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.

Input

The first line contains one integer n (2≤n≤100000)n (2≤n≤100000).

The second line contains nn integers, a1,a2,…,ana1,a2,…,an (1≤ai≤2000001≤ai≤200000).

Output

Print one integer: gcd({lcm({ai,aj}) | i<j})gcd({lcm({ai,aj}) | i<j}).

Examples

input

Copy

2
1 1

output

Copy

1

input

Copy

4
10 24 40 80

output

Copy

40

input

Copy

10
540 648 810 648 720 540 594 864 972 648

output

Copy

54

Note

For the first example, t={lcm({1,1})}={1}t={lcm({1,1})}={1}, so gcd(t)=1gcd(t)=1.

For the second example, t={120,40,80,120,240,80}t={120,40,80,120,240,80}, and it's not hard to see that gcd(t)=40gcd(t)=40.

 题意：一个长度为 N 的数组，求 gcd {lcm({ai , aj}) | i < j}

  gcd_1=gcd( lcm(a1 , a2) , lcm(a1 , a3) ... lcm(a1 , an) ) 

  gcd_2=gcd( lcm(a2 , a3) , lcm(a2, a4) ... lcm(a2 , an) )

 ……

  gcd_n=gcd( lcm(an , 0))

因为gcd_1一定是a1的倍数，所以：

 gcd_1=lcm (a1 , gcd (a2 , a3 , ... an) )

#include <bits/stdc++.h>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const int mod = 998244353;
const int N = 3e5+5;
inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m%n);
}
ll lcm(ll m, ll n)
{
    return m*n / gcd(m, n);
}
bool cmp(PII a, PII b)
{
    return a.second<b.second;
}
int a[N],suf[N];
int main() {
    int n,ans=0;
    cin>>n;
    rep(i,1,n)
    {
        cin>>a[i];
    }
    dec(i,n,1)
    {
        suf[i]=gcd(suf[i+1],a[i]);
    }
    rep(i,1,n)
    {
        ans=gcd(ans,lcm(suf[i+1],a[i]));
    }
    cout<<ans<<endl;
    return 0;
}