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.io.*;
import java.math.*;

public class practice {
    public static void main(String[] args) throws Exception {
            Scanner in = new Scanner(System.in);
            int p = in.nextInt();
            int n = in.nextInt();
            boolean[] arr = new boolean[p];
            boolean good = false;
            for(int i = 0; i < n; i++) {
                long l = in.nextLong();
                if(!arr[(int) l % p]) {
                arr[(int) l % p] = true;
                } else {
                    System.out.println(i + 1);
                    good = true;
                    break;
                }
            }
            if(!good) System.out.println(-1);
        /*StringBuilder sb = new StringBuilder();
        for (int i=0; i<a; i++){
            sb.append(nums[i]);
            if(i != a - 1) {
                sb.append(' ');
            }
        }
        System.out.println(sb.toString());*/
    }
}