#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<int,P> P1;

#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define rep(i,x) for(int i=0;i<x;i++)
#define rep1(i,x) for(int i=1;i<=x;i++)
#define rrep(i,x) for(int i=x-1;i>=0;i--)
#define rrep1(i,x) for(int i=x;i>0;i--)
#define sor(v) sort(v.begin(),v.end())
#define rev(s) reverse(s.begin(),s.end())
#define lb(vec,a) lower_bound(vec.begin(),vec.end(),a)
#define ub(vec,a) upper_bound(vec.begin(),vec.end(),a)
#define uniq(vec) vec.erase(unique(vec.begin(),vec.end()),vec.end())
#define mp1(a,b,c) P1(a,P(b,c))

const int INF=1000000000;
const int dir_4[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
const int dir_8[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};

const long long M = 998244353;

long long gcd(long long x,long long y){
	if(y == 0)return x;
	return gcd(y,x%y);
}

int main(){
	long long N;
	long long a[100010];
	scanf("%lld",&N);
	for(int i = 1 ; i <= N ; i ++){
		scanf("%lld",&a[i]);
	}
	sort(a+1,a+N+1);
	
	long long ret = 1;
	for(int i = 1 ; i <= N ; i ++){
		ret *= gcd(a[i],i-1);
		ret %= M;
	}
	cout << ret << endl;
}

### Prompt

Your task is to create a CPP solution to 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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<int,P> P1;

#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define rep(i,x) for(int i=0;i<x;i++)
#define rep1(i,x) for(int i=1;i<=x;i++)
#define rrep(i,x) for(int i=x-1;i>=0;i--)
#define rrep1(i,x) for(int i=x;i>0;i--)
#define sor(v) sort(v.begin(),v.end())
#define rev(s) reverse(s.begin(),s.end())
#define lb(vec,a) lower_bound(vec.begin(),vec.end(),a)
#define ub(vec,a) upper_bound(vec.begin(),vec.end(),a)
#define uniq(vec) vec.erase(unique(vec.begin(),vec.end()),vec.end())
#define mp1(a,b,c) P1(a,P(b,c))

const int INF=1000000000;
const int dir_4[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
const int dir_8[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};

const long long M = 998244353;

long long gcd(long long x,long long y){
	if(y == 0)return x;
	return gcd(y,x%y);
}

int main(){
	long long N;
	long long a[100010];
	scanf("%lld",&N);
	for(int i = 1 ; i <= N ; i ++){
		scanf("%lld",&a[i]);
	}
	sort(a+1,a+N+1);
	
	long long ret = 1;
	for(int i = 1 ; i <= N ; i ++){
		ret *= gcd(a[i],i-1);
		ret %= M;
	}
	cout << ret << endl;
}

```