// includes {{{
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<random>
#include<cassert>
// #include<deque>
// #include<multiset>
// #include<bitset>
// #include<cstring>
// #include<bits/stdc++.h>
// }}}
using namespace std;
using ll = long long;

/// --- math {{{ ///
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll extgcd(ll a, ll b, ll &x, ll &y) {
  ll d;
  return b == 0 ? (x = 1, y = 0, a) : (d = extgcd(b, a % b, y, x), y -= a / b * x, d);
}
ll modinv(ll a, ll mod) {
  ll x, y;
  extgcd(a, mod, x, y);
  if(x < 0) x += mod;
  return x;
}
ll modpow(ll a, ll b, ll mod) {
  ll r = 1;
  a %= mod;
  while(b) {
    if(b & 1) r = r * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
/// }}}--- ///

const int mod = 998244353;

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  int a[112345];
  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 = ans * gcd(a[i], i) % mod;
  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
// includes {{{
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<random>
#include<cassert>
// #include<deque>
// #include<multiset>
// #include<bitset>
// #include<cstring>
// #include<bits/stdc++.h>
// }}}
using namespace std;
using ll = long long;

/// --- math {{{ ///
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll extgcd(ll a, ll b, ll &x, ll &y) {
  ll d;
  return b == 0 ? (x = 1, y = 0, a) : (d = extgcd(b, a % b, y, x), y -= a / b * x, d);
}
ll modinv(ll a, ll mod) {
  ll x, y;
  extgcd(a, mod, x, y);
  if(x < 0) x += mod;
  return x;
}
ll modpow(ll a, ll b, ll mod) {
  ll r = 1;
  a %= mod;
  while(b) {
    if(b & 1) r = r * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
/// }}}--- ///

const int mod = 998244353;

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  int a[112345];
  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 = ans * gcd(a[i], i) % mod;
  cout << ans << endl;
  return 0;
}

```