#include <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <bitset>
#include <complex>
#include <unistd.h>
#include <cassert>
#include <cctype>
#include <random>
#define _USE_MATH_DEFINES
#define _GLIBCXX_DEBUG
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> plglg;
typedef pair<double, ll> pdlg;
typedef tuple<int, int, int> tiii;
typedef tuple<ll, ll, ll> tlglglg;
typedef tuple<double, double, double> tddd;
typedef complex<double> xy_t;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
double pi = 3.141592653589793;
ll mod = 998244353;
int intmax = 2147483647;
int intmin = -2147483648;
ll llmax = 9223372036854775807;
ll llmin = -9223372036854775807;
int iinf = intmax / 8;
ll inf = llmax / 8;
double eps = 1e-11;

ll multiply(ll x, ll y) {
    return (x % mod) * (y % mod) % mod;
}

ll gcd(ll a, ll b) {
    return b != 0 ? gcd(b, a % b) : a;
}

int main() {
    ll N;
    cin >> N;
    ll a[N];
    for (int i = 0; i < N; i++) {
        cin >> a[i];
    }
    sort(a, a + N);
    ll ans = 1;
    for (int i = 0; i < N; i++) {
        ans = multiply(ans, gcd(a[i], i));
    }
    cout << ans << 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 <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <bitset>
#include <complex>
#include <unistd.h>
#include <cassert>
#include <cctype>
#include <random>
#define _USE_MATH_DEFINES
#define _GLIBCXX_DEBUG
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> plglg;
typedef pair<double, ll> pdlg;
typedef tuple<int, int, int> tiii;
typedef tuple<ll, ll, ll> tlglglg;
typedef tuple<double, double, double> tddd;
typedef complex<double> xy_t;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
double pi = 3.141592653589793;
ll mod = 998244353;
int intmax = 2147483647;
int intmin = -2147483648;
ll llmax = 9223372036854775807;
ll llmin = -9223372036854775807;
int iinf = intmax / 8;
ll inf = llmax / 8;
double eps = 1e-11;

ll multiply(ll x, ll y) {
    return (x % mod) * (y % mod) % mod;
}

ll gcd(ll a, ll b) {
    return b != 0 ? gcd(b, a % b) : a;
}

int main() {
    ll N;
    cin >> N;
    ll a[N];
    for (int i = 0; i < N; i++) {
        cin >> a[i];
    }
    sort(a, a + N);
    ll ans = 1;
    for (int i = 0; i < N; i++) {
        ans = multiply(ans, gcd(a[i], i));
    }
    cout << ans << endl;
}

```