#include<bits/stdc++.h>
using namespace std;
#define maxn 100020
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define inf 0x3f3f3f3f
typedef long long ll;
typedef pair<int,int> pr;

const ll mod = 998244353;
ll ans = 1;
int a[maxn],n;

ll gcd(ll x,ll y){ 
	if ( !y ) return x;
	return gcd(y,x % y);
}
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",&a[i]);
	sort(a + 1,a + n + 1);
	rep(i,1,n) ans = ans * gcd(a[i],i - 1) % mod;
	cout<<ans<<endl;
}

### Prompt

In Cpp, your task is to solve the following problem:

Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).

\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.

Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.

The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.

Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).

Constraints

* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers

Input

Input is given from Standard Input in the following format:


\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)


Output

Print the answer.

Examples

Input

2
4 3


Output

3


Input

4
2 5 2 5


Output

2

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define maxn 100020
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define inf 0x3f3f3f3f
typedef long long ll;
typedef pair<int,int> pr;

const ll mod = 998244353;
ll ans = 1;
int a[maxn],n;

ll gcd(ll x,ll y){ 
	if ( !y ) return x;
	return gcd(y,x % y);
}
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",&a[i]);
	sort(a + 1,a + n + 1);
	rep(i,1,n) ans = ans * gcd(a[i],i - 1) % mod;
	cout<<ans<<endl;
}

```