Niwango has N cards, numbered 1,2,\ldots,N. He will now arrange these cards in a row.

Niwango wants to know if there is a way to arrange the cards while satisfying all the N conditions below. To help him, determine whether such a way exists. If the answer is yes, also find the lexicographically smallest such arrangement.

* To the immediate right of Card 1 (if any) is NOT Card a_1.
* To the immediate right of Card 2 (if any) is NOT Card a_2.
* \vdots
* To the immediate right of Card N (if any) is NOT Card a_N.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq a_i \leq N
* a_i \neq i

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 \ldots a_N


Output

If no arrangements satisfy the conditions, print `-1`. If such arrangements exist, print the lexicographically smallest such arrangement, in the following format:


b_1 b_2 \ldots b_N


Here, b_i represents the i-th card from the left.

Examples

Input

4
2 3 4 1


Output

1 3 2 4


Input

2
2 1


Output

-1


Input

13
2 3 4 5 6 7 8 9 10 11 12 13 12


Output

1 3 2 4 6 5 7 9 8 10 12 11 13
//Zory-2020
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define Debug printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define debug(x) cerr<<#x<<'='<<x<<endl
#define mem(x,val) memset(x,val,sizeof x)
namespace mine
{
    ll qread()
    {
        ll ans=0,f=1;char c=getchar();
        while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
        while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
        return ans*f;
    }
    void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
    void write1(ll num){write(num);putchar(' ');}
    void write2(ll num){write(num);putchar('\n');}
    template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
    template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
    ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
    bool IN(ll x,ll l,ll r){return l<=x and x<=r;}

    const int INF=0x3f3f3f3f;
    const int MOD=1e9+7;
    int mm(const int x){return x>=MOD?x-MOD:x;}
    template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
    ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
    ll invm(ll x){return qpower(x,MOD-2);}
    const int N=1e5+10;

    set<int> a;typedef set<int>::iterator IT;
    int n,b[N],ban[N],deg[N];
    bool dfs(int lst)
    {
        if(!sz(a)) return 1;
        // fo(i,1,n-sz(a)) write1(b[i]);puts("");
        for(IT it=a.begin();it!=a.end();)
        {
            int nxt=*it;if(nxt==ban[lst]) {++it;continue;}
            b[n-sz(a)+1]=nxt;a.erase(nxt);deg[ban[nxt]]--;
            if((sz(a)-1!=deg[ban[nxt]] or !a.count(ban[nxt])) and dfs(nxt)) return 1;
            it=a.upper_bound(nxt);a.insert(nxt);deg[ban[nxt]]++;
        }return 0;
    }
    void main()
	{
        n=qread();fo(i,1,n) a.insert(i),deg[ban[i]=qread()]++;
        if(!dfs(0)) puts("-1"); else fo(i,1,n) write1(b[i]);
	}
};//(ans+MOD)%MOD
signed main()
{
    srand(time(0));
    mine::main();
}
