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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define bit(n) (1<<(n))
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<bool> vb;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
int const INF = 1001001001;


// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
long long val;
constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
if (val < 0) val += MOD;
}
constexpr int getmod() { return MOD; }
constexpr Fp operator - () const noexcept {
return val ? MOD - val : 0;
}
constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
constexpr Fp& operator += (const Fp& r) noexcept {
val += r.val;
if (val >= MOD) val -= MOD;
return *this;
}
constexpr Fp& operator -= (const Fp& r) noexcept {
val -= r.val;
if (val < 0) val += MOD;
return *this;
}
constexpr Fp& operator *= (const Fp& r) noexcept {
val = val * r.val % MOD;
return *this;
}
constexpr Fp& operator /= (const Fp& r) noexcept {
long long a = r.val, b = MOD, u = 1, v = 0;
while (b) {
long long t = a / b;
a -= t * b; swap(a, b);
u -= t * v; swap(u, v);
}
val = val * u % MOD;
if (val < 0) val += MOD;
return *this;
}
constexpr bool operator == (const Fp& r) const noexcept {
return this->val == r.val;
}
constexpr bool operator != (const Fp& r) const noexcept {
return this->val != r.val;
}
friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
return os << x.val;
}
friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
if (n == 0) return 1;
auto t = modpow(a, n / 2);
t = t * t;
if (n & 1) t = t * a;
return t;
}
};

const int MOD = 1000000007;
using mint = Fp<MOD>;

// main() {vi is_prime = Eratos(MAX);}
// is_prime[i]:= 1 or 0
vector<int> Eratos(int MAX) {
    vector<int> is_prime(MAX,1);
    is_prime[0] = 0; is_prime[1] = 0;
    for(int i = 2; i < MAX; ++i) {
        if(!is_prime[i]) continue;
        for(int j = i*2; j < MAX; j += i) is_prime[j] = 0;
    }
    return is_prime;
}

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

vi P = Eratos(1000);
map<int, int> mp;
rrep(i,999) {
    if(P[i]) mp[i] = 0;
}

int N; cin >> N;
vi A(N);
rep(i,N) {
    cin >> A[i];
}
rep(i,N) {
    int cur = A[i];
    for(auto x : mp) {
        if(cur == 1) break;
        int cnt = 0;
        while(cur % x.fi == 0) {
            cur /= x.fi;
            cnt++;
        }
        mp[x.fi] = max(x.se, cnt);
    }
    if(cur != 1) mp[cur] = 1;
}

mint L = 1;
for(auto x : mp) {
    L *= modpow((mint)x.fi, x.se);
}
mint sum = 0;
rep(i,N) {
    sum += L / A[i];
}
cout << sum << endl;
return 0;
}