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.*;

//							------ SRM 425 ------
/*An out-of-control robot is placed on a plane, and it takes n random steps. 
 * At each step, the robot chooses one of four directions randomly and moves 
 * one unit in that direction. The probabilities of the robot choosing north,
 *  south, east or west are north, south, east and west percent, respectively.
 * The robot's path is considered simple if it never visits the same point more 
 * than once. (The robot's starting point is always the first visited point.) 
 * Return a double containing the probability that the robot's path is simple. 
 * For example, "EENE" and "ENW" are simple, but "ENWS" and "WWWWSNE" are not 
 * ('E', 'W', 'N' and 'S' represent east, west, north and south, respectively).
 */
public class Ejemplo{

	public static void main(String args[]){
		Scanner scan = new Scanner(System.in);
		int p = scan.nextInt();
		int n = scan.nextInt();
		int [] v = new int [301];
		int ans = 0;
		for(int i = 0; i < 301; i++){
			v[i] = -1;
		}
		boolean f = true;
		for(int i = 0; i < n; i++){
			int tem = scan.nextInt();
			if (v[tem % p] == -1){
				v[tem % p] = tem;
			}
			else{
				if(f){
					ans = i;
					f = false;
				}
			}
		}
		if(ans == 0){
			System.out.println(-1);
		}
		else{
			System.out.println(ans+1);
		}
	}
}
/*
test 1:
	14
	50
	50
	0
	0
Returns: 1.220703125E-4
Here, the only possible directions are east and west. The only two available paths are
"EEEEEEEEEEEEEE" and "WWWWWWWWWWWWWW". The probability is equal to 2 / (2^14).
 */
