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 Codeforce_447_A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan = new Scanner(System.in);
		int nBuckets=scan.nextInt(), nNumbers=scan.nextInt();
		int arrFilledBuckets[]=new int[nBuckets];
		int nFBuckets=0, result=-1;
		for (int i=0; i<nNumbers;i++)
		{
			int x=scan.nextInt();
			x=x%nBuckets;
			//System.out.println(x);
			if ((nFBuckets==0)||(search(x,arrFilledBuckets,nFBuckets)==false))
			{
				arrFilledBuckets[nFBuckets]=x;
				nFBuckets++;			
			}
			else
			{
				if(result==-1)result=i+1;
			}
		}
		System.out.println(result);
		scan.close();
	}

	private static boolean search(int i, int[] arrFilledBuckets, int nFBuckets) {
		// TODO Auto-generated method stub
		for(int j=0; j<nFBuckets; j++)if(i==arrFilledBuckets[j])return true;
		return false;
	}

}
