Greedy Change

Greedy Change

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Billy investigates the question of applying greedy algorithm to different spheres of life. At the moment he is studying the application of greedy algorithm to the problem about change. There is an amount of n coins of different face values, and the coins of each value are not limited in number. The task is to collect the sum x with the minimum amount of coins. Greedy algorithm with each its step takes the coin of the highest face value, not exceeding x. Obviously, if among the coins' face values exists the face value 1, any sum x can be collected with the help of greedy algorithm. However, greedy algorithm does not always give the optimal representation of the sum, i.e. the representation with the minimum amount of coins. For example, if there are face values {1, 3, 4} and it is asked to collect the sum6, greedy algorithm will represent the sum as 4 + 1 + 1, while the optimal representation is 3 + 3, containing one coin less. By the given set of face values find out if there exist such a sum x that greedy algorithm will collect in a non-optimal way. If such a sum exists, find out the smallest of these sums.

Input

The first line contains an integer n (1 ≤ n ≤ 400) — the amount of the coins' face values. The second line contains n integers ai(1 ≤ ai ≤ 109), describing the face values. It is guaranteed that a1 > a2 > ... > an and an = 1.

Output

If greedy algorithm collects any sum in an optimal way, output -1. Otherwise output the smallest sum that greedy algorithm collects in a non-optimal way.

Examples

input

5
25 10 5 2 1

output

-1

input

3
4 3 1

output

6
分析：据说是论文结论题。
　　　A Polynomial-time Algorithm for the Change-Making Problem；
　　　由结论，这个数比a[i]大一点点;
　　　所以先贪心a[i]-1，然后枚举j(j>i),把a[j]的数目+1，然后再贪心，看是不是数目变大了；
　　　如果变大了，则取一个最小的答案；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<ll,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
const int maxn=4e2+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,ans,a[maxn],b[maxn],c[maxn];
bool flag;
int solve(int p,int *q)
{
    int ret=0;
    for(int i=1;i<=n;i++)
    {
        q[i]=p/a[i];
        p%=a[i];
        ret+=q[i];
    }
    return ret;
}
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,n)scanf("%d",&a[i]);
    rep(i,1,n)
    {
        solve(a[i]-1,b);
        rep(j,i+1,n)
        {
            int now=0,num=0;
            rep(k,1,j-1)now+=a[k]*b[k],num+=b[k];
            now+=a[j]*(b[j]+1),num+=b[k]+1;
            if(solve(now,c)>num)
            {
                if(!flag)
                {
                    flag=true;
                    ans=now;
                }
                else ans=min(ans,now);
            }
        }
    }
    if(flag)printf("%d
",ans);
    else puts("-1");
    //system("Pause");
    return 0;
}