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.util.Scanner;

public class DZYLovesHash {
	public static void main(String[] args) {
		 System.out.println(findConflictNumber(getNumbers()));
	}

	private static int[] getNumbers() {

		Scanner s = new Scanner(System.in);
		int p = s.nextInt();
		int n = s.nextInt();

		int[] number = new int[n+1];
		for (int i = 0; i < n; i++)
			number[i] = s.nextInt();
		s.close();
		
		number[n] = p;
		return number;
	}

	private static int findConflictNumber(int[] numbers) {
		int conflict_number = -1;
		int p = numbers[(numbers.length-1)];
		int[] mod = new int[p];
		java.util.Arrays.fill(mod, -1);
		for (int i = 0; i < (numbers.length-1) ; i++) {
			
			// to find the remained value of dividing the "given number" to "p"
			int remained = numbers[i]%(p);

			if (mod[remained] == -1)
				mod[remained] = remained;
			else
				return i+1;
		}
		return conflict_number;
	}
}
