#include<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<int32_t mod>
struct ModInt{
	int32_t a;
	inline int32_t normalize(int64_t x){
		x%=mod;
		if(x<0)x+=mod;
		return x;
	}
	ModInt(const int64_t a=0):a(normalize(a)){}

	ModInt& operator+=(const ModInt &x){
		a+=x.a;
		if(a>=mod)a-=mod;
		return *this;
	}
	ModInt& operator-=(const ModInt &x){
		a-=x.a;
		if(a<0)a+=mod;
		return *this;
	}
	ModInt& operator*=(const ModInt &x){
		a=(int64_t)a*x.a%mod;
		return *this;
	}
	ModInt& operator/=(const ModInt &x){
		*this*=x.inv();
		return *this;
	}

	ModInt operator+(const ModInt x){return ModInt(*this)+=x;}
	ModInt operator-(const ModInt x){return ModInt(*this)-=x;}
	ModInt operator*(const ModInt x){return ModInt(*this)*=x;}
	ModInt operator/(const ModInt x){return ModInt(*this)/=x;}
	bool operator==(const ModInt x){return a==x.a;}
	bool operator!=(const ModInt x){return a!=x.a;}

	ModInt operator-(){return ModInt(0)-ModInt(*this);}
	ModInt pow(int64_t ex)const{
		int64_t x=a;
		int64_t res=1;
		while(ex){
			if(ex&1)res=res*x%mod;
			x=x*x%mod;
			ex>>=1;
		}
		return ModInt(res);
	}
	ModInt inv()const{return pow(mod-2);}
};

template<int32_t mod>
istream& operator>>(istream& in,ModInt<mod>& a){
	return (in>>a.a);
}
template<int32_t mod>
ostream& operator<<(ostream& out,const ModInt<mod>& a){
	return (out<<a.a);
}
using mint=ModInt<998244353>;

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

int N;
int A[111111];

signed main(){
	cin>>N;
	rep(i,N)cin>>A[i];
	sort(A,A+N);

	mint ans=1;
	for(int i=0;i<N;i++)ans*=gcd(A[i],i);
	cout<<ans<<endl;
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<int32_t mod>
struct ModInt{
	int32_t a;
	inline int32_t normalize(int64_t x){
		x%=mod;
		if(x<0)x+=mod;
		return x;
	}
	ModInt(const int64_t a=0):a(normalize(a)){}

	ModInt& operator+=(const ModInt &x){
		a+=x.a;
		if(a>=mod)a-=mod;
		return *this;
	}
	ModInt& operator-=(const ModInt &x){
		a-=x.a;
		if(a<0)a+=mod;
		return *this;
	}
	ModInt& operator*=(const ModInt &x){
		a=(int64_t)a*x.a%mod;
		return *this;
	}
	ModInt& operator/=(const ModInt &x){
		*this*=x.inv();
		return *this;
	}

	ModInt operator+(const ModInt x){return ModInt(*this)+=x;}
	ModInt operator-(const ModInt x){return ModInt(*this)-=x;}
	ModInt operator*(const ModInt x){return ModInt(*this)*=x;}
	ModInt operator/(const ModInt x){return ModInt(*this)/=x;}
	bool operator==(const ModInt x){return a==x.a;}
	bool operator!=(const ModInt x){return a!=x.a;}

	ModInt operator-(){return ModInt(0)-ModInt(*this);}
	ModInt pow(int64_t ex)const{
		int64_t x=a;
		int64_t res=1;
		while(ex){
			if(ex&1)res=res*x%mod;
			x=x*x%mod;
			ex>>=1;
		}
		return ModInt(res);
	}
	ModInt inv()const{return pow(mod-2);}
};

template<int32_t mod>
istream& operator>>(istream& in,ModInt<mod>& a){
	return (in>>a.a);
}
template<int32_t mod>
ostream& operator<<(ostream& out,const ModInt<mod>& a){
	return (out<<a.a);
}
using mint=ModInt<998244353>;

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

int N;
int A[111111];

signed main(){
	cin>>N;
	rep(i,N)cin>>A[i];
	sort(A,A+N);

	mint ans=1;
	for(int i=0;i<N;i++)ans*=gcd(A[i],i);
	cout<<ans<<endl;
	return 0;
}
```