#include <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 998244353;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////

int main() {
	ll N;
	cin >> N;
	vll a(N + 1);
	repn(i, N) cin >> a[i];
	sort(++a.begin(), a.end());

	mint ans = a[1];
	repn(i, N - 1) {
		ans *= gcd(a[i + 1], i);
	}
	cout << ans << endl;
}
### Prompt

Develop a solution in Cpp 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 <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 998244353;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////

int main() {
	ll N;
	cin >> N;
	vll a(N + 1);
	repn(i, N) cin >> a[i];
	sort(++a.begin(), a.end());

	mint ans = a[1];
	repn(i, N - 1) {
		ans *= gcd(a[i + 1], i);
	}
	cout << ans << endl;
}
```