Problem:
Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int mod = 1000000007;

#define REP(i, n)  for(int i = 0; i < n; ++i)
#define FOR(i, t, n)  for(ll i = t; i <= (ll)n; ++i)

struct mint {
    ll x; // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}

    mint operator-() const { return mint(-x); }

    mint &operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }

    mint &operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }

    mint &operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }

    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const {
        return pow(mod - 2);
    }

    mint &operator/=(const mint a) {
        return (*this) *= a.inv();
    }

    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};


class Eratosthenes {
public:
    vector<bool> prime;

    Eratosthenes(ll size) {
        prime.resize(size, 1);
        prime[0] = 0;
        prime[1] = 0;
        for (size_t i = 2; i * i <= size; ++i)
            if (prime[i]) {
                for (size_t j = 2; i * j <= size; ++j)prime[i * j] = false;
            }
    }
};


class Combination {
    vector<ll> fac; //n!(mod M)
    vector<ll> ifac; //k!^{M-2} (mod M)
public:
    Combination(ll size) {
        fac.resize(size);
        ifac.resize(size);
        fac[0] = 1;
        ifac[0] = 1;
        for (ll i = 0; i < size; i++) {
            fac[i + 1] = fac[i] * (i + 1) % mod;
            ifac[i + 1] = ifac[i] * mpow(i + 1, mod - 2) % mod;
        }
    }

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

    ll comb(ll a, ll b) {
        if (a == 0 && b == 0)return 1;
        if (a < b || a < 0)return 0;
        ll tmp = ifac[a - b] * ifac[b] % mod;
        return tmp * fac[a] % mod;
    }
};


ll GCD(ll a, ll b) {
    if (b == 0)return a;
    return GCD(b, a % b);
}

ll LCM(ll a, ll b) {
    ll gcd = GCD(a, b);
    return a / gcd * b;
}

string s;
ll N;

int main() {
    cin >> N;
    Eratosthenes e(1e6 + 1e5);
    map<int, int> m;
    vector<ll> A(N);
    REP(k, N) {
        cin >> A[k];
        ll a = A[k];
        for (int i = 2; i * i <= e.prime.size(); i++) {
            if (!e.prime[i])continue;
            int cnt = 0;
            while (a % i == 0)a /= i, cnt++;
            m[i] = max(m[i], cnt);
        }
        if (a > 1)m[a] = max(m[a], 1);
    }

    mint lcm = 1;
    for (auto k : m) REP(i, k.second)lcm *= k.first;

    mint sum = 0;
    REP(i, N) {
        mint x = lcm / A[i];
        sum += x;
    }
    cout << sum.x << endl;

    return 0;
}