CF 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

//题意就是找相等的数，输出第二个的位置，可是要是最先发现的。
比如：10 5
      1 2 2 2 1
输出是3而不是5，由于先找到2和2相等，假设仅仅用for循环，找到的是1和1相等输出是5.
第4个例子卡了非常久，没看懂题目。
。。。

#include <iostream>
using namespace std;
int main()
{   __int64 a[400];
    int n,t,i,j,p,k;
    while(scanf("%d%d",&p,&n)!=EOF)
    {   memset(a,0,sizeof(a));
        t=0;
        for(i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
            a[i]=a[i]%p;
        }
        k=n;
        for(i=0;i<n-1;i++)
        {    
            for(j=i+1;j<n;j++)
                if(a[i]==a[j])
                {   
                    k=k<(j+1)?
k:(j+1);
                        t=1;
                }
                
        }
        if(t==1)
            printf("%d
",k);
       if(t==0)
           printf("-1
");
    }
    return 0;
}