#include<algorithm>
#include<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<fstream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

template < typename T >
void vprint(T &V){
	for(auto v : V){
    	cout << v << " ";
	}
	cout << endl;
}

ll N;
vector<ll> A;

bool is_kadomatsu(ll i, ll j, ll k){
    if(A[i]==A[j] || A[j]==A[k]) return false;
    if(A[i]<A[j] && A[j]<A[k]) return false;
    if(A[i]>A[j] && A[j]>A[k]) return false;
    return true;
}

ll memo[1<<12];

// win : flag
// lose : -1 
ll can_win(ll f){
    if(memo[f]!=0) return memo[f];
    
    FOR(i, 0, N){
        FOR(j, i+1, N){
            FOR(k, j+1, N){
                if(f>>i & 1 || f>>j & 1 || f>>k & 1) continue; // 使用済みをもう1度使おうとした場合
                if(!is_kadomatsu(i, j, k)) continue;

                ll f2 = (1<<i) + (1<<j) + (1<<k); // 今回選んだフラグ
                ll f3 = f + f2;

                // 相手が勝てない手を押し付けられる
                if(can_win(f3)==-1){
                    memo[f] = f2;
                    return memo[f];
                }
            }
        }
    }

    // 押し付けられなかった
    memo[f] = -1;
    return memo[f];
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    cin >> N;
    A.resize(N);

    FOR(i, 0, N){
        cin >> A.at(i);
    }

    ll f = can_win(0);
    if(f==-1){
        p(-1);
    }else{
        vector<ll> answer;
        FOR(i, 0, 12){
            if(f>>i & 1){
                answer.push_back(i);
            }
        }
        
        cout << answer[0] << " " << answer[1] << " " << answer[2] << endl;
    }

    return 0;
}