[CodeForces - 447A] A - DZY Loves Hash

A - DZY Loves Hash

DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert nnumbers, 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 5021534153

Output

4

Input

5 501234

Output

-1

题目的意思是,按顺序给出几个数,判断从哪个操作开始,hash值有重复.

题目很水,模拟一下就好了.

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 ];
 int main(){
     int n,m;
     scanf("%d%d",&m,&n);
     ; i<=n; i++){
         int x; scanf("%d",&x);
         x%=m; ;}
         vis[x]=;
     }
     puts("-1");
     ;
 }