#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;
typedef long long  ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
typedef vector<int> vec;

int gcd(int a, int b) {
	if (a < b)swap(a, b);
	while (b) {
		int r = a % b; a = b; b = r;
	}
	return a;
}

void solve() {
	//init();

	int n; cin >> n;
	vector<int> a(n);
	rep(i, n)cin >> a[i];
	sort(a.begin(), a.end());
	ll ans = 1;
	rep(i, n) {
		ll g = gcd(i, a[i]);
		ans = ans * g%mod;
	}
	cout << ans << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(5);
	//while (cin >> n,n)solve();
	solve();
	//stop
	return 0;
}

### 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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;
typedef long long  ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
typedef vector<int> vec;

int gcd(int a, int b) {
	if (a < b)swap(a, b);
	while (b) {
		int r = a % b; a = b; b = r;
	}
	return a;
}

void solve() {
	//init();

	int n; cin >> n;
	vector<int> a(n);
	rep(i, n)cin >> a[i];
	sort(a.begin(), a.end());
	ll ans = 1;
	rep(i, n) {
		ll g = gcd(i, a[i]);
		ans = ans * g%mod;
	}
	cout << ans << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(5);
	//while (cin >> n,n)solve();
	solve();
	//stop
	return 0;
}

```