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

public class DZYLovesHash {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        
        int P = sc.nextInt();
        int N = sc.nextInt();
        
        int[] input = new int[N];
        for (int i = 0; i < N; i++) {
        	input[i] = sc.nextInt();
        }
        
        HashSet<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < N; i++) {
        	int x = input[i];
        	int h = x % P;
        	if (set.contains(h)) {
        		System.out.println(i + 1);
        		return;
        	}
        	set.add(h);
        }
        System.out.println(-1);
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            String str = "";
            try { str = br.readLine(); }
            catch (IOException e) { e.printStackTrace(); }
            return str;
        }
    }
}