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.*;
import java.math.*;
import java.io.*;
import java.awt.Point;

import static java.util.Arrays.*;
import static java.lang.Integer.*;
import static java.lang.Double.*;
import static java.lang.Long.*;
import static java.lang.Short.*;
import static java.lang.Math.*;
import static java.math.BigInteger.*;
import static java.util.Collections.*;

public class A {
	public static void main(String[] xa) {
		try (Scanner sc = new Scanner(new BufferedInputStream(System.in));
				PrintWriter out = new PrintWriter(new BufferedWriter(
						new OutputStreamWriter(System.out)))) {
			int p = sc.nextInt();
			int n = sc.nextInt();
			boolean[] x = new boolean[p];
			int a = -1;
			int ins = 0;
			while (n-- > 0) {
				int s = sc.nextInt() % p;
				ins++;
				if (a == -1) {
					if (x[s]) {
						a = ins;
					}
					x[s] = true;
				}
			}
			out.println(a);
			out.flush();
			out.close();
		}
	}
}