C. Orac and LCM

链接：https://codeforces.ml/contest/1350/problem/C

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

input

Copy

4
10 24 40 80

output

Copy

input

Copy

10
540 648 810 648 720 540 594 864 972 648

output

Copy

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.

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<map>
#include<algorithm>
using namespace std;
#define ll long long
#define lb long double
#define INF 0x3f3f3f3f
#define maxn 200010
ll n,k,l,t,x,s;
ll a[200001];
ll dp[100001];
map<ll,ll>m;
int main()
{cin>>n;for(int i=1;i<=n;i++){cin>>a[i];m[a[i]]++;}k=1;for(ll i=2;i<=200000;i++){for(ll j=i+i;j<=200000;j+=i){m[i]+=m[j];}if(m[i]>=n-1){k*=i/__gcd(i,k);}}cout<<k;
}

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