题意：

A. Worms Evolution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are nforms of worms. Worms of these forms have lengths a1, a2, ..., an. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.

Input

The first line contains integer n (3 ≤ n ≤ 100) — amount of worm's forms. The second line contains n space-separated integers ai(1 ≤ ai ≤ 1000) — lengths of worms of each form.

Output

Output 3 distinct integers i j k (1 ≤ i, j, k ≤ n) — such indexes of worm's forms that ai = aj + ak. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that aj = ak.

Examples

input

Copy

5
1 2 3 5 7

output

Copy

3 2 1

input

Copy

5
1 8 1 5 1

output

Copy

-1

解析：

输入n个数字，判断数组里面是否存在两个数字相加等于数组里面的另外的一个数字并且返回其下标

代码：

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    int a[1000];
    cin>>n;
    if (n>=3&&n<=100)
    {
        
    
    for (int x=1; x<=n; x++)
    {
        cin>>a[x];
    }
    for (int x=1; x<=n; x++)
    {
        for (int y=1; y<=n; y++)
        {
            for (int z=1;z<=n ; z++)
            {
                if (a[z]==a[y]+a[x]&&x!=y&&x!=z&&z!=y)
                {
                    cout<<z<<" "<<y<<" "<<x;
                        return 0;
                    
                }
            }
        }
    }
    cout<<-1;
    }
    return  0;
}