Problem:
Score : 500 points

Problem Statement

An adult game master and N children are playing a game on an ice rink.
The game consists of K rounds.
In the i-th round, the game master announces:

Form groups consisting of A_i children each!

Then the children who are still in the game form as many groups of A_i children as possible.
One child may belong to at most one group.
Those who are left without a group leave the game. The others proceed to the next round.
Note that it's possible that nobody leaves the game in some round.

In the end, after the K-th round, there are exactly two children left, and they are declared the winners.

You have heard the values of A_1, A_2, ..., A_K. You don't know N, but you want to estimate it.

Find the smallest and the largest possible number of children in the game before the start, or determine that no valid values of N exist.

Constraints

1 \leq K \leq 10^5

2 \leq A_i \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

K
A_1 A_2 ... A_K

Output

Print two integers representing the smallest and the largest possible value of N, respectively,
or a single integer -1 if the described situation is impossible.

Sample Input 1

4
3 4 3 2

Sample Output 1

6 8

For example, if the game starts with 6 children, then it proceeds as follows:

In the first round, 6 children form 2 groups of 3 children, and nobody leaves the game.

In the second round, 6 children form 1 group of 4 children, and 2 children leave the game.

In the third round, 4 children form 1 group of 3 children, and 1 child leaves the game.

In the fourth round, 3 children form 1 group of 2 children, and 1 child leaves the game.

The last 2 children are declared the winners.

Sample Input 2

5
3 4 100 3 2

Sample Output 2

-1

This situation is impossible.
In particular, if the game starts with less than 100 children, everyone leaves after the third round.

Sample Input 3

10
2 2 2 2 2 2 2 2 2 2

Sample Output 3

2 3

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

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <utility>
#include <tuple>
#include <cctype>
using namespace std;
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3fLL
#define MOD 1000000007
#define mp make_pair
#define mt make_tuple
#define pb push_back
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pint;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> tint;
typedef vector<int> vint;
typedef vector<ll> vll;
int dx[8]={0,0,-1,1,1,1,-1,-1};
int dy[8]={-1,1,0,0,1,-1,1,-1};
const int SIZE=1050;
//ここまでテンプレ
int main(){
	int K;
	cin>>K;
	vll A;
	for(int i=0;i<K;i++){
		ll a;
		cin>>a;
		A.pb(a);
	}
	reverse(A.begin(),A.end());
	for(int i=1;i<K;i++){
		if(A[i]>=2*A[i-1]){
			cout<<-1<<endl;
			return 0;
		}
	}
	ll mini=2,maxi=2;
	for(int i=0;i<K;i++){
		ll now=A[i];
		//miniは、現時点でのmini以上で最小のnowの倍数
		if(mini%now)
			mini=mini-(mini%now)+now;
		//maxiは、現時点でのmaxi以上で最小のnowの倍数
		if(maxi%now)
			maxi=maxi-(maxi%now)+now;
	}
	cout<<mini<<" "<<maxi+A.back()-1<<endl;
	return 0;
}