Team Olympiad

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此文章探讨了Berland首都一所学校如何从擅长不同科目的学生中组建最多数量的科学十项全能奥林匹克竞赛队伍。每支队伍由一名擅长编程、一名擅长数学及一名擅长体育的学生组成。通过算法确定了最大队伍数并给出了具体组建方案。

Description

The School №0 of the capital of Berland has n children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value t_i:

t_i = 1, if the i-th child is good at programming,
t_i = 2, if the i-th child is good at maths,
t_i = 3, if the i-th child is good at PE

Each child happens to be good at exactly one of these three subjects.

The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.

What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?

Input

The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t₁, t₂, ..., t_n (1 ≤ t_i ≤ 3), where t_i describes the skill of the i-th child.

Output

In the first line output integer w — the largest possible number of teams.

Then print w lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to n in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.

If no teams can be compiled, print the only line with value w equal to 0.

Sample Input

Input

7
1 3 1 3 2 1 2

Output

2
3 5 2
6 7 4

Input

4
2 1 1 2

Output

#include<cstdio>
#include<iostream>
using namespace std;

int num1[5010],num2[5010],num3[5010];

int main(){
    int n,i,t,sum,one,two,three,zuixiao;
    while(cin>>n){
        one=two=three=sum=0;
        for(i=1;i<=n;i++){
            cin>>t;
            if(t==1){
                num1[one]=i;
                one++;
            }
            else if(t==2){
                num2[two]=i;
                two++;
            }
            else if(t==3){
                num3[three]=i;
                three++;
            }
        }
        zuixiao=min(one,min(two,three));
        cout<<zuixiao<<endl;
        for(i=0;i<zuixiao;i++){
            cout<<num1[i]<<" "<<num2[i]<<" "<<num3[i]<<endl;
        }
    }
    return 0;
}