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 thei-th number x_i, DZY will put it into the bucket numberedh(x_i), whereh(x) is the hash function. In this problem we will assume, thath(x) = x mod p. Operationa mod b denotes taking a remainder after divisiona 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 thei-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. Thei-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 5021534153

Output

4

Input

5 501234

Output

-1

//题意就是找相等的数，输出第二个的位置，但是要是最先发现的。例如：10 5      1 2 2 2 1输出是3而不是5，因为先找到2和2相等，如果只用for循环，找到的是1和1相等输出是5.第4个样例卡了很久，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\n",k);       if(t==0)           printf("-1\n");    }    return 0;}