codeforces471B

MUH and Important Things

 CodeForces - 471B 

It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are ntasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.

Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the n tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of tasks. The second line contains n integers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 2000), where h_i is the difficulty of the i-th task. The larger number h_i is, the more difficult the i-th task is.

Output

In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line ndistinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.

If there are multiple possible answers, you can print any of them.

Examples

Input

4
1 3 3 1

Output

YES
1 4 2 3 
4 1 2 3 
4 1 3 2 

Input

5
2 4 1 4 8

Output

NO

Note

In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.

In the second sample there are only two sequences of tasks that meet the conditions — [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.

sol：XJB构造三串不同的字典序最小的序列，十分容易，两个相同的就两两交换，多个相同的就用第一个与第二个或第三个（最后一个）交换，这样就凑到三种情况了

Ps：可能构造多种会比较困难，感觉只会n！的方法（GG）

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=2005;
int n,Hash[N];
struct data
{
    int Shuz;
}a[N];
int Pailie[N];
int Used[N];
vector<int>Jih[N];
int main()
{
    int i,j,Fas=1;
    R(n);
    for(i=1;i<=n;i++)
    {
        R(a[i].Shuz); Hash[++*Hash]=a[i].Shuz;
    }
    sort(Hash+1,Hash+*Hash+1);
    *Hash=unique(Hash+1,Hash+*Hash+1)-Hash-1;
    for(i=1;i<=n;i++)
    {
        a[i].Shuz=lower_bound(Hash+1,Hash+*Hash+1,a[i].Shuz)-Hash;
        Jih[a[i].Shuz].push_back(i);
    }
    for(i=1;i<=*Hash&&Fas<3;i++)
    {
        if(Jih[i].size()==2) Fas*=2;
        else if(Jih[i].size()>2) Fas=3;
    }
    if(Fas<3) return 0*puts("NO");
    puts("YES");
    for(int Step=1;Step<=3;Step++)
    {
        bool Bo=0;
        for(i=1;i<=*Hash;i++)
        {
            if(Bo)
            {
                Used[i]=1;
                for(j=0;j<Jih[i].size();j++) W(Jih[i][j]);
                continue;
            }
            if(Jih[i].size()==1)
            {
                W(Jih[i][0]); continue;    
            }
            else if(Jih[i].size()==2)
            {
                if(Used[i]==0)
                {
                    W(Jih[i][0]); W(Jih[i][1]); Bo=1;
                    Used[i]++;
                }
                else if(Used[i]==1)
                {
                    W(Jih[i][1]); W(Jih[i][0]); Bo=1;
                    Used[i]++;
                }
                else
                {
                    W(Jih[i][0]); W(Jih[i][1]);
                }
            }
            else
            {
                if(Used[i]==0)
                {
                    for(j=0;j<Jih[i].size();j++) W(Jih[i][j]);
                    Bo=1;
                    Used[i]++;
                }
                else if(Used[i]==1)
                {
                    W(Jih[i][1]); W(Jih[i][0]);
                    for(j=2;j<Jih[i].size();j++) W(Jih[i][j]);
                    Bo=1;
                    Used[i]++;
                }
                else
                {
                    W(Jih[i][Jih[i].size()-1]);
                    for(j=1;j<Jih[i].size()-1;j++) W(Jih[i][j]);
                    W(Jih[i][0]);
                    Bo=1;
                }
            }
        }
        puts("");
    }
    return 0;
}
/*
input
4
1 3 3 1
output
YES
1 4 2 3 
4 1 2 3 
4 1 3 2 
*/