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)

/*

#      #      #####   ##   ##  #####      ##     #  #      #   #  #####
#      #      #   #  #  ###  # #          # ##   #  #      #   #  #   #
#      #      #   #  #   #   # #          #  ##  #  #      #   #  #   #
#      #      #   #   #     #  #####      #   #  #  #       ###   #####
#      #      #   #    #   #   #          #  ##  #  #        #    #   #
#      #      #   #     ###    #          # ##   #  #        #    #   #
#      #####  #####      #     #####      ##     #  #####    #    #   #

*/

#include <bits/stdc++.h>

using namespace std;

#define F first
#define S second
#define pb push_back
#define np next_permutation
#define bs binary_search
#define mp make_pair
#define all(x) x.begin(), x.end()
#define bp __builtin_popcountll
#define sqr(x) ((llong)sqrt (x))
#define b_ctz __builtin_ctzll
#define b_clz __builtin_clzll

#define y1 road_to_respa_top_6
#define prev road_to_respa_top_6
#define Abdrakhman abcdefghijklmnopqrstuvwxyzRakhman
#define left kotoriy_levee_chem_right
#define right kotoriy_pravee_chem_left

#define android ios_base::sync_with_stdio(0), cin.tie (0), cout.tie (0)
#define fn ""

typedef long long llong;
typedef long double ldouble;
typedef unsigned long long ullong;

typedef vector <llong> vl;
typedef vector <string> vs;
typedef vector <vector <llong>> matrix;

typedef pair <llong, llong> pll;

typedef map <llong, llong> mll;

typedef set <llong> sl;

const llong MXN = 1e6 + 1;
const llong MNN = 1e3 + 1;
const llong MOD = 1e9 + 7;
const llong INF = 1e18;
const llong MAGIC = 1000;
const ldouble EPS = 1e-9;
const ldouble PI = 3.1415926536;
const string TIME = "rovno_ushten_push_ketti";

inline llong log2 (llong x) {return 64 - b_clz (x) - 1;};
inline llong __lcm (llong a, llong b) {return a / __gcd (a, b) * b;};

llong n;
llong a[MXN];

llong get (llong x) {
    for (llong i = 1; i <= n; ++ i) {
        x -= x % a[i];
    }
    return x;
}

int main () {
    android;
    random_device rand;
    #ifdef I_LOVE_DILYA
        freopen ("input.txt", "r", stdin);
    #endif
    cin >> n;
    for (llong i = 1; i <= n; ++ i) cin >> a[i];
    llong x, y;
    llong l = 0, r = MOD;
    while (r - l > 1) {
        llong mid = (l + r) / 2;
        if (get (mid) <= 2) l = mid;
        else r = mid;
    }
    if (get (r) <= 2) x = r;
    else x = l;
    l = 0, r = MOD;
    while (r - l > 1) {
        llong mid = (l + r) / 2;
        if (2 <= get (mid)) r = mid;
        else l = mid;
    }
    if (2 <= get (l)) y = l;
    else y = r;
    if (x < y) return cout << "-1", 0;
    cout << y << ' ' << x;
    return 0;
}