Problem: Petya has got 2n cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2n. We'll denote the number that is written on a card with number i, as ai. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.

Input

The first line contains integer n (1 ≤ n ≤ 3·105). The second line contains the sequence of 2n positive integers a1, a2, ..., a2n (1 ≤ ai ≤ 5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.

Output

If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print n pairs of integers, a pair per line — the indices of the cards that form the pairs.

Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.

Examples

Input

3
20 30 10 30 20 10


Output

4 2
1 5
6 3


Input

1
1 2


Output

-1

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Hashtable;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.StringTokenizer;

public class A
{
    public static void main(String[] args) throws IOException
    {
        PrintWriter out = new PrintWriter(new File("output.txt"));
//       PrintWriter out = new PrintWriter(System.out);
        InputA in = new InputA();
        int n = in.nextInt();
        ArrayList<ArrayList<Integer>> L = new ArrayList<ArrayList<Integer>>(5001);
        for(int i = 0 ; i < 5001 ; i++)
            L.add(new ArrayList<Integer>());
        for (int i = 1; i <= 2 * n; i++)
            L.get(in.nextInt()).add(i);
        boolean valid = true;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < L.size() && valid; i++)
        {
            if (L.get(i).size() % 2 == 0)
                for (int j = 0; j + 1 < L.get(i).size(); j += 2)
                    sb.append(L.get(i).get(j) + " " + L.get(i).get(j + 1) + "\n");
            else
                valid = false;
        }
        if (!valid)
            out.println(-1);
        else
            out.print(sb.toString());
        out.close();
    }
}

class InputA
{
    BufferedReader bf;
    StringTokenizer st;

    public InputA() throws IOException
    {
//       bf = new BufferedReader(new InputStreamReader(System.in));
        bf = new BufferedReader(new FileReader(new File("input.txt")));
        st = new StringTokenizer(bf.readLine());
    }

    public String next() throws IOException
    {
        while (!st.hasMoreTokens() && bf.ready())
        {
            st = new StringTokenizer(bf.readLine());
        }
        return st.hasMoreTokens() ? st.nextToken() : null;
    }

    public boolean hasNext() throws IOException
    {
        while (!st.hasMoreTokens() && bf.ready())
        {
            st = new StringTokenizer(bf.readLine());
        }
        return st.hasMoreTokens();
    }

    public int nextInt() throws IOException
    {
        while (!st.hasMoreTokens() && bf.ready())
        {
            st = new StringTokenizer(bf.readLine());
        }
        return st.hasMoreTokens() ? new Integer(st.nextToken()) : 0;

    }

    public long nextLong() throws IOException
    {
        while (!st.hasMoreTokens() && bf.ready())
        {
            st = new StringTokenizer(bf.readLine());
        }
        return st.hasMoreTokens() ? new Long(st.nextToken()) : 0l;

    }

}