// TLE解，O(N^2)

#include <bits/stdc++.h>

using namespace std;
#define all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {os << "["; for (const auto &v : vec) {os << v << ","; } os << "]"; return os;}
template <typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &p) {os << "(" << p.first << ", " << p.second << ")"; return os;}

const int mod = 1e9 + 7;

ll mod_pow(ll x, ll n, ll mod) {
    ll res = 1;
    while(n > 0) {
        if(n & 1) res = res * x % mod;
            x = x * x % mod;
            n >>= 1;
    }
    return res;
}

ll inv(ll x) {
    return mod_pow(x, mod - 2, mod);
}

ll f(ll x, ll y) {
    return (((x + y) % mod) * mod_pow(x, y, mod)) % mod;
}

// (x + y)x^y < (s + t)s^t
// log(x + y) + ylog(x) < log(s + t) + tlog(s)
double g(ll x, ll y) {
    return log(x + y) + y * log(x);
}

void solve() {
    int N;
    cin >> N;
    vector<ll> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
        A[i] %= mod;
    }
    //sort(all(A));

    ll prod = 1;

    int ii = 0, jj = 1;
    double mi = g(A[0], A[1]);
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
            prod *= f(A[i], A[j]);
            prod %= mod;
            if (mi > g(A[i], A[j])) {
                mi = g(A[i], A[j]);
                ii = i, jj = j;
            }
        }
    }
    cout << (prod * inv(f(A[ii], A[jj]))) % mod << endl;
    
}


int main() {
    #ifdef LOCAL_ENV
    cin.exceptions(ios::failbit);
    #endif
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    
    solve();
}