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>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define rrep(i,n) for (ll i = (n)-1; i >=0; --i)
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

using namespace std;
using ll = long long;
using ld = long double;
using pi = pair<ll, ll>;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const ll mod = 1000000007;
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 { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= 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 { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} c(200005);



int main() {
  ll k;
  cin >> k;

  vector<ll> v(k);
  rep(i, k) cin >> v[i];

  reverse(v.begin(), v.end());
  v.push_back(1);
  
  ll dp[k+1][2];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  dp[0][1] = 1;

  rep(i, k){
    // cerr << i << endl;
    // cerr << v[i] << " " << dp[i][0]  << " " << dp[i][1] <<endl;

    // cerr << v[i]*dp[i][0]/v[i+1] << " " << v[i]*(dp[i][0]+1) << endl;
    // for(int j=(v[i]*dp[i][0])/v[i+1]; j*v[i+1]<v[i]*(dp[i][0]+1); j++){
    //   if(j*v[i+1]>=max(v[i]*dp[i][0], v[i+1])){
    // 	dp[i+1][0] = j;
    // 	cerr << "#" << endl;
    // 	break;
    //   }
    // }
    dp[i+1][0] = (v[i]*dp[i][0]+v[i+1]-1)/v[i+1];
      cerr << dp[i+1][0]*v[i+1] << " " << (dp[i][0]+1)*v[i] << endl;
    // if(dp[i+1][0]*v[i+1]>(dp[i][0]+1)*v[i] || dp[i+1][0]==0){
    //   cout << -1 << endl;
    //   return 0;
    //}
      
  }
    // for(int j=v[i]*dp[i][0]; j<v[i]*(dp[i][0]+1); j++){
    //   if(j>=v[i+1] && j%v[i+1]==0) {
    // 	dp[i+1][0] = j/v[i+1]; 
    // 	break;
    //   }
    // }

  rep(i, k){
    // cerr << i << endl;
    // cerr << v[i] << " " << dp[i][0]  << " " << dp[i][1] <<endl;

    // cerr << v[i]*dp[i][0]/v[i+1] << " " << v[i]*(dp[i][0]+1) << endl;
    for(int j=v[i]*(dp[i][1]+1)/v[i+1]+1; j*v[i+1]>=v[i]*(dp[i][0]); j--){
      if(j*v[i+1]>=v[i+1] && j*v[i+1]<v[i]*(dp[i][1]+1)){
	dp[i+1][1] = j;
	break;
      }
    }
  }
  if(dp[0][0]*v[0]!=2){
    cout << -1 << endl;
    return 0;
  }
  
  cout << v[k-1] *dp[k-1][0] << " "<<v[k-1] *(dp[k-1][1]+1)-1 <<  endl;
  return 0;
}