D - DZY Loves Hash CodeForces - 447A

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.

Example

Input

10 5
0
21
53
41
53

Output

4

Input

5 5
0
1
2
3
4

 题意：把n个数字放到0~p-1个位置，要求x放在x%p的位置；如果已经放过 输出x的位置，否则输出-1；
（吐槽一下这个题，一直wa就是找不到原因，最后发现原来是没有数组清零，真是太坑了；）
代码：

#include<iostream>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
using namespace std;

int main(){
   int n,p,k,flag=;
   //char c[301][20];
   char m[];

   unsigned long long x;
   cin>>p>>n;
   for(int i = ;i<p;i++)
   {
       m[i]=;
   }//数组一定要清零，
   for(int i=;i<n;i++){
    cin>>x;
    k=x%p;
    if(m[k]==)
        m[k]=;
    else if(!flag)
        flag=i+;
   }
   if(flag)cout<<flag<<endl;
   else cout<<-<<endl;

}

 Output

-1