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 Sample {

    public static void main(String[] args) {
        int i,j;
        Scanner sc = new Scanner(System.in);
        int n,m,k,p;
        p = sc.nextInt();
        n = sc.nextInt();
        int[] a = new int[n];
        for(i=0;i<n;i++) {
            a[i] = sc.nextInt();
        }
        for (i=0;i<n;i++) {
            a[i] = a[i]%p;
        }
        for (i=1;i<n;i++) {
            for (j=0;j<i;j++) {
                if (a[i]==a[j]) {
                    System.out.println(i+1);
                    return;
                }
            }
        }
        System.out.println("-1");
    }
}







