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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define MOD 1000000007
#define INF (1LL<<57)
#define MV 2000001
#define LMV 21
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define emp emplace
#define whoami(x) cerr<<#x<<" "<<x<<"\n";
#define mp make_pair
#define ins insert
#define sz size

void FLASH() {ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);}
void SETF() {cout.ios_base::setf(ios_base::fixed);}
void UNSETF() {cout.ios_base::unsetf(ios_base::fixed);}

typedef long long ll;
typedef long double ld;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<PII, int> PPII;
typedef pair<PLL, ll> PPLL;
typedef map<int, int> MII;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

struct h_llint {
  static uint64_t splitmix64(uint64_t x) {    // http://xorshift.di.unimi.it/splitmix64.c
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};

struct h_pair{
  size_t operator()(const PLL&x)const{
    return hash<ll>()(((ll)x.ff)^(((ll)x.ss)<<32));
  }
};

typedef map<ll, ll> MLL;
typedef map<PII, int> MPII;
typedef map<PLL, ll> MPLL;
typedef set<int> SI;
typedef set<ll> SL;
//ordered_set = order_by_key(.)
//ordered_set = find_by_order(.)
typedef ordered_set<int> OSI;
typedef ordered_set<ll> OSL;
typedef ordered_multiset<int> MOSI;
typedef ordered_multiset<ll> MOSL;
typedef unordered_map<ll, int, h_llint> UMLI;
typedef unordered_map<ll, ll, h_llint> UMLL;
typedef unordered_map<PLL, int, h_pair> UMPI;
typedef unordered_map<PLL, ll, h_pair> UMPL;

int ar[MV];
ll lo[MV], arr[MV];

ll modexp(ll a, ll n)
{
	ll res = 1;
	while(n)
	{
		if(n & 1)
		res = ((res%MOD) * (a%MOD))%MOD;
		a = ((a%MOD) * (a%MOD))%MOD;
		n >>= 1;	
	}
	return res;
}

void siever(void)
{
	iota(lo, lo+MV, 0);
	for(ll p=2;p*p<MV;p++)
	{
		if(lo[p] == p)
		{
			for(ll i=p*p;i<MV;i+=p)
			{
				if(lo[i] == i)
				lo[i] = p;	
			}
		}
	}
	return;
}

void solve(int T)
{
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
	cin>>arr[i];
	ll ax = 1;
	for(int i=0;i<n;i++)
	{
		ll tx = arr[i];
		while(tx > 1)
		{
			int ct = 0;
			ll p = lo[tx];
			while(!(tx % p))
			{
				tx /= p;
				ct++;
			}	
			ar[p] = max(ar[p], ct);
		}
	}
	ll sx = 0;
	for(int i=0;i<MV;i++)
	ax = ((ax%MOD) * (modexp(i, ar[i])%MOD))%MOD;
	for(int i=0;i<n;i++)
	sx = ((sx%MOD) + (((ax%MOD) * (modexp(arr[i], MOD-2)%MOD))%MOD))%MOD;
	cout<<sx<<"\n";
	return;
}

int main(void)
{
	FLASH();
	int T;
	T = 1;
	siever();

// #ifndef ONLINE_JUDGE
// 	time_t time_t1, time_t2;
// 	time_t1 = clock();
// #endif

	while(T--)
	solve(T);

// #ifndef ONLINE_JUDGE
// 	time_t2 = clock();
// 	SETF();
// 	cout<<"Time taken: "<<setprecision(7)<<(time_t2 - time_t1)/(double)CLOCKS_PER_SEC<<"\n";
// 	UNSETF();
// #endif

	return 0;
}