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// Marcin Knapik

#pragma GCC optimize ("O3")
#include<bits/stdc++.h>
using namespace std;

#define FOR(i, n) for (int i = 0; i < n; i++)
#define f first
#define s second
#define pb push_back
#define all(s) s.begin(), s.end()
#define sz(s) (int)s.size()

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;

template <class T>ostream &operator<<(ostream &os, vector<T> &vec){for (T &el : vec){os << el << ' ';}return os;}
template <class T>istream &operator>>(istream &is, vector<T> &vec) {for (T &el : vec){is >> el;}return is;}

template <class T, class G> ostream &operator<<(ostream &os, pair<T, G> para) { os << para.f << ' ' << para.s; return os;}

using ll = long long;
using vi = vector<int>;
using ii = pair<int, int>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vll = vector<ll>;
using vvpll = vector<vpll>;
using vii = vector<ii>;
using mp = map<vector<string>, long double>;
using ld = long double;

#define rep(i,a,b) for(int i = a; i < b; i++)
const ll mod = 1e9 + 7;
ll det(vector<vector<ll>>& a) {
	int n = sz(a); ll ans = 1;
	rep(i,0,n) {
		rep(j,i+1,n) {
			while (a[j][i] != 0) { // gcd step
				ll t = a[i][i] / a[j][i];
				if (t) rep(k,i,n)
					a[i][k] = (a[i][k] - a[j][k] * t) % mod;
				swap(a[i], a[j]);
				ans *= -1;
			}
		}
		ans = ans * a[i][i] % mod;
		if (!ans) return 0;
	}
	return (ans + mod) % mod;
}

void solve(){
    int n;
    cin >> n;

    vi tab(n);
    for(int i = 0; i < n; i++)
        cin >> tab[i];
    
    vector<vector<ll>> pom(n - 1, vll(n - 1));
    
    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            int GCD = __gcd(tab[i], tab[j]);
            if(i < n - 1)
                pom[i][i] += GCD;
            if(j < n - 1)
                pom[j][j] += GCD;
            
            if(i < n - 1 and j < n - 1){
                pom[i][j] -= GCD;
                pom[j][i] -= GCD;
            }
        }
    }

    cout << det(pom) << '\n';
}

int main () {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int tests = 1;
    // cin >> tests;

    for (int test = 1; test <= tests; test++) {
        solve();
    }

    return 0;
}