#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <array>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <fstream>
#include <complex>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
using namespace std;

using ll = long long;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

struct mint {
    int x;
    mint() : x(0) {}
    mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    mint& operator+=(const mint& p){
        if((x += p.x) >= mod)   x -= mod;
        return *this;
    }
    mint& operator-=(const mint& p){
        if((x -= p.x) < 0)  x += mod;
        return *this;
    }
    mint& operator*=(const mint& p){
        x = (int)(1LL * x * p.x % mod);
        return *this;
    }
    mint& operator/=(const mint& p){
        *this *= p.inverse();
        return *this;
    }
    mint operator-() const { return mint(-x); }
    mint operator+(const mint& p) const { return mint(*this) += p; }
    mint operator-(const mint& p) const { return mint(*this) -= p; }
    mint operator*(const mint& p) const { return mint(*this) *= p; }
    mint operator/(const mint& p) const { return mint(*this) /= p; }
    bool operator==(const mint& p) const { return x == p.x; }
    bool operator!=(const mint& p) const { return x != p.x; }
    mint pow(int64_t n) const {
        mint res = 1, mul = x;
        while(n > 0){
            if(n & 1)   res *= mul;
            mul *= mul;
            n >>= 1;
        }
        return res;
    }
    // x^(a^b)
    // warning : x と mod は互いに素
    // x % mod == 0 かつ a > 0 はこれを呼び出さずに 0 を返すように処理
    mint pow2(int64_t a, int64_t b) const {
        if(b == 0)  return *this;
        if((a %= mod - 1) == 0) return mint(1);
        int64_t n = 1;
        while(b > 0){
            if(b & 1)   (n *= a) %= mod - 1;
            (a *= a) %= mod - 1;
            b >>= 1;
        }
        return pow(n);
    }
    mint inverse() const { return pow(mod - 2); }
    friend ostream& operator<<(ostream& os, const mint& p){
        return os << p.x;
    }
    friend istream& operator>>(istream& is, mint& p){
        int64_t val;
        is >> val;
        p = mint(val);
        return is;
    }
};

mint dp[1000005];

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

    int K, S = 0;
    cin >> K;
    vector<int> C(K);
    for(int i = 0; i < K; ++i){
        cin >> C[i];
        S += C[i];
    }

    vector<int> div;
    for(int i = 1; i * i <= S; ++i){
        if(S % i == 0){
            div.emplace_back(i);
            if(i * i != S)  div.emplace_back(S / i);
        }
    }
    sort(begin(div), end(div));

    vector<mint> fact(S + 1, 1), ifact(S + 1, 1);
    for(int i = 1; i <= S; ++i) fact[i] = fact[i - 1] * i;
    ifact[S] = fact[S].inverse();
    for(int i = S; i > 0; --i)  ifact[i - 1] = ifact[i] * i;

    int N = div.size();
    mint ans = 0;
    for(int i = 0; i < N; ++i){
        if(div[i] < K)  continue;
        
        bool ok = true;
        int sum = 0, cnt = S / div[i];
        mint way = fact[div[i]];
        for(int j = 0; j < K; ++j){
            if(C[j] % cnt != 0) { ok = false; break; }
            int per = C[j] / cnt;
            sum += per;
            way *= ifact[per];
        }
        ok &= sum == div[i];
        if(!ok) continue;

        dp[div[i]] += way;
        ans += dp[div[i]] / div[i];
        for(int j = i + 1; j < N; ++j){
            if(div[j] % div[i] == 0) dp[div[j]] -= dp[div[i]];
        }
    }

    cout << ans << endl;

    return 0;
}