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 S447A {

	public int solve(int p, int n, int[] h) {
		boolean[] b = new boolean[p];
		
		for (int i = 0; i < n; i ++) {
			int j = h[i] % p;
			if (!b[j])
				b[j] = true;
			else
				return i + 1;
		}
		
		return -1;
	}
	public static void main(String[] args) {
		
		S447A a = new S447A();
		Scanner in = new Scanner(System.in);
		
		int p = in.nextInt();
		int n = in.nextInt();
		
		int[] h = new int[n];
		for (int i = 0; i < n; i ++)
			h[i] = in.nextInt();
		
		System.out.println(a.solve(p, n, h));
		
		
		/*int p0 = 10;
		int n0 = 5;
		int[] h0 = {0,21,53,41,53};
		
		int p1 = 5;
		int n1 = 5;
		int[] h1 = {0,1,2,3,4};
		
		System.out.println(a.solve(p0, n0, h0));
		System.out.println(a.solve(p1, n1, h1));
		*/
		
	}

}
