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
#include<set>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N;
int A[100011];
int C[100011];
bool B[100011];

bool valid(const VI &v) {
    REP (i, N-1) if (A[v[i]] == v[i+1]) return false;
    return true;
}

void MAIN() {
    scanf("%d", &N);
    REP (i, N) scanf("%d", A+i), A[i]--;

    REP (i, N) C[A[i]]++;
    set<pair<int, int> > FI;
    set<int> se;
    REP (i, N) {
	FI.emplace(C[i], i);
	se.insert(i);
    }

    VI guess;
    REP (t, N) {
	int rest = N-t;
	if (rest <= 4) {
	    VI tmp = guess;
	    VI v;
	    REP (j, N) if (!B[j]) v.push_back(j);
	    do {
		REP (j, v.size()) tmp.push_back(v[j]);

		if (valid(tmp)) {

		    EACH (e, tmp) (*e)++;
		    rprintf("%d", tmp.begin(), tmp.end());
		    return;
		}

		REP (j, v.size()) tmp.pop_back();
	    } while (next_permutation(v.begin(), v.end()));
	    break;
	}
	int k;
	if (FI.rbegin()->first == rest-1) {
	    k = FI.rbegin()->second;
	} else {
	    k = *se.begin();
	    if (t && A[guess.back()] == k) {
		k = *next(se.begin());
	    }
	}

	guess.push_back(k);
	FI.erase(make_pair(C[k], k));
	B[k] = true;
	se.erase(k);
	if (!B[A[k]]) {
	    FI.erase(make_pair(C[A[k]], A[k]));
	    C[A[k]]--;
	    FI.emplace(C[A[k]], A[k]);
	}
    }
    EACH (e, guess) (*e)++;
    puts("-1");
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

