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.*;
import java.util.*;

public class A implements Runnable {

	private void Solution() throws IOException {
		int n = nextInt();
		ArrayList<Integer>[] a = new ArrayList[5001];
		for (int i = 0; i < 5001; i++)
			a[i] = new ArrayList<Integer>();
		for (int i = 1; i <= 2 * n; i++) {
			int cur = nextInt();
			a[cur].add(i);
		}
		boolean ok = true;
		for (int i = 0; i < 5001; i++)
			if (a[i].size() % 2 != 0)
				ok = false;
		if (ok) {
			for (int i = 0; i < 5001; i++)
				for (int j = 0; j < a[i].size(); j += 2)
					println(a[i].get(j), a[i].get(j + 1));
		} else
			println(-1);
	}

	public static void main(String[] args) {
		new A().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer tokenizer;

	public void run() {
		try {
			long beginTime = System.nanoTime();
			long beginMemory = totalMemory() - freeMemory();

			String file = "";
			file = "";
			if (file.equals("")) {
				in = new BufferedReader(new FileReader(new File("input.txt")));
				out = new PrintWriter(new File("output.txt"));
			} else {
				in = new BufferedReader(new FileReader(new File(file + ".in")));
				out = new PrintWriter(new File(file + ".out"));
			}
			Solution();
			out.close();

			long endTime = System.nanoTime();
			long endMemory = totalMemory() - freeMemory();

			double totalTime = (endTime - beginTime) * 1d / 1000000000;
			double totalMemory = (endMemory - beginMemory) * 1d / 1024 / 1024;

			System.err.println("Time wasted - " + totalTime + " s");
			System.err.println("Memory used - " + totalMemory + " mb");
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(261);
		}
	}

	long totalMemory() {
		return Runtime.getRuntime().totalMemory();
	}

	long freeMemory() {
		return Runtime.getRuntime().freeMemory();
	}

	void halt() {
		out.close();
		System.exit(0);
	}

	void print(Object... obj) {
		for (int i = 0; i < obj.length; i++) {
			if (i != 0)
				out.print(" ");
			out.print(obj[i]);
		}
	}

	void println(Object... obj) {
		print(obj);
		out.println();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

	String nextLine() throws IOException {
		return in.readLine();
	}

	String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(nextLine());
		return tokenizer.nextToken();
	}
}