DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number xi, DZY will put it into the bucket numbered h(xi), where h(x) is the hash function. In this problem we will assume, that h(x) = x mod p. Operation a mod b denotes taking a remainder after division a by b.

However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the i-th insertion, you should output i. If no conflict happens, just output -1.

Input

The first line contains two integers, p and n (2 ≤ p, n ≤ 300). Then n lines follow. The i-th of them contains an integer xi (0 ≤ xi ≤ 109).

Output

Output a single integer — the answer to the problem.

Examples

Input

10 5
0
21
53
41
53


Output

4


Input

5 5
0
1
2
3
4


Output

-1
import java.io.*;
import java.util.*;

public class round300 {
	static int p, n;
	static int[] nums;
	static boolean[] arr;
	public static void main(String args[]) {
		Scanner in;
		try {
			in = new Scanner(new File("input.txt"));
		}
		catch (Exception e) {
			in = new Scanner(System.in);
		}
		p = in.nextInt();
		n = in.nextInt();
		nums = new int[n];
		arr = new boolean[p];
		for (int i=0; i<p; i++) arr[i]=false;
		for (int i=0; i<n; i++) {
			nums[i]=in.nextInt();
		}
		System.out.println(cool());
	}
	public static int cool() {
		for (int i = 0; i < nums.length; i++) {
			int ex = nums[i] % p;
			if (arr[ex]) return i+1;
			arr[ex]=true;
		}
		return -1;
	}
}