CF A. DZY Loves Hash

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.

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 a single integer ― the answer to the problem.

10 5
0
21
53
41
53

4

5 5
0
1
2
3
4

-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 
   
    
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