/* --- author: dxm --- */
#include<bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;

void qread(int &x){
	int neg=1;x=0;
	char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')neg=-1;c=getchar();}
	while(c>='0'&&c<='9')x=10*x+c-'0',c=getchar();
	x*=neg;
}

const int maxn=100005;
const int mod=998244353;
int n,ans=1;
int a[maxn];

int gcd(int a,int b){
	return !b?a:gcd(b,a%b);
}

int main(){
#ifdef FILE
	freopen("FILE.in","r",stdin);
	freopen("FILE.out","w",stdout);
#endif
	scanf("%d",&n);
	rep(i,n)scanf("%d",&a[i]);
	sort(a,a+n);
	rep(i,n)ans=1LL*ans*gcd(a[i],i)%mod;
	printf("%d\n",ans);
#ifdef TIME
	printf("Running Time = %d ms\n",int(clock()*1000.0/CLOCKS_PER_SEC));
#endif
	return 0;
}

/*
Input:
-----------------
Output:
*/

### Prompt

Construct a Cpp code solution to the problem outlined:

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
/* --- author: dxm --- */
#include<bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;

void qread(int &x){
	int neg=1;x=0;
	char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')neg=-1;c=getchar();}
	while(c>='0'&&c<='9')x=10*x+c-'0',c=getchar();
	x*=neg;
}

const int maxn=100005;
const int mod=998244353;
int n,ans=1;
int a[maxn];

int gcd(int a,int b){
	return !b?a:gcd(b,a%b);
}

int main(){
#ifdef FILE
	freopen("FILE.in","r",stdin);
	freopen("FILE.out","w",stdout);
#endif
	scanf("%d",&n);
	rep(i,n)scanf("%d",&a[i]);
	sort(a,a+n);
	rep(i,n)ans=1LL*ans*gcd(a[i],i)%mod;
	printf("%d\n",ans);
#ifdef TIME
	printf("Running Time = %d ms\n",int(clock()*1000.0/CLOCKS_PER_SEC));
#endif
	return 0;
}

/*
Input:
-----------------
Output:
*/

```