Problem:
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Score : 1000 points

Problem Statement

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.

Sample Input 1

2
4 3

Sample Output 1

3

In this case, \(N = 2, a = [ 4, 3 ]\).

For the permutation \(p = [ 1, 2 ]\), a cycle of an only chicken \(1\) and a cycle of an only chicken \(2\) are made, thus \(f([1, 2]) = a_1 \times a_2 = 12\).

For the permutation \(p = [ 2, 1 ]\), a cycle of two chickens \(1\) and \(2\) is made, and the size of the smallest chicken is \(a_2 = 3\), thus \(f([2, 1]) = a_2 = 3\).

Now we know \(b_1 = f([2, 1]) = 3, b_2 = f([1, 2]) = 12\), and the greatest common divisor of \(b_1\) and \(b_2\) is \(3\).

Sample Input 2

4
2 5 2 5

Sample Output 2

2

There are always \(N!\) permutations because chickens of the same size can be distinguished from each other.

The following picture illustrates the cycles formed and their beauties when \(p = (2, 1, 4, 3)\) and \(p = (3, 4, 1, 2)\), respectively.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#define int long long
#define p 99824435

int gcd(int a, int b){
	if(b == 0){
		return a;
	}
	else{
		return gcd(b, a % b);
	}
}

signed compare(const void *a, const void *b){
	int sub = *(int *)a - *(int *)b;
	if(sub < 0){
		return -1;
	}
	else if(sub == 0){
		return 0;
	}
	else{
		return 1;
	}
}

signed main(){
	int N, i, ans = 1;
	scanf("%lld", &N);
	int *A = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%lld", &A[i]);
	}
	qsort(A, N, sizeof(int), compare);
	for(i = 0; i < N; i++){
		ans = (ans * gcd(A[i], i)) % p;
	}
	printf("%lld\n", ans);
	return 0;
}