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
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

/**
 *
 * @author RezaM
 */
public class PA {

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stt = new StringTokenizer(br.readLine());

        int p = Integer.parseInt(stt.nextToken());
        int n = Integer.parseInt(stt.nextToken());

        boolean bary[] = new boolean[p];

        int i;

        for (i = 0; i < n; i++) {

            int x = Integer.parseInt(br.readLine());
            int index = x % p;

            if (!bary[index]) {
                bary[index] = true;
            } else {
                break;
            }

        }

        if (i != n) {
            System.out.println(i+1);
        } else {
            System.out.println(-1);
        }

    }

}
