Codeforces Round #FF (Div. 2)A. DZY Loves Hash



A. DZY Loves Hash

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x_i, DZY will put it into the bucket numbered h(x_i), 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 x_i (0 ≤ x_i ≤ 10⁹).

Output

Output a single integer — the answer to the problem.

Sample test(s)

Input

10 5
0
21
53
41
53

Output

4

Input

5 5
0
1
2
3
4

Output

-1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,p;
int a[310],count1;
int main()
{
	int num;
	while(scanf("%d%d",&p,&n)!=EOF)
	{
		int flag=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&num);
			a[i]=num%p;
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<i;j++)
			{
				if(a[j]==a[i])
				{
					flag=0;
					count1=i;
					break;
				}
			}
			if(!flag) break;
		}
		if(!flag) cout<<count1<<endl;
		else cout<<"-1"<<endl;
	}
	return 0;
}