Codeforces Round #FF (Div. 2) A.DYZ Loves Hash

水題。題目大意爲輸入一個p,n，再輸入n組數據，每組數據對p取餘，當和前面相同時，發現衝突，記下第幾組，找到最後一組沒有找到和前面一樣的輸出-1。

*可以採用一個標記數組，就變得很簡單了。

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 xi, DZY will put it into the bucket numbered h(xi), 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 xi (0 ≤ xi ≤ 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

#include<stdio.h>
#include<string.h>
int main()
{
    int p,n;
    while(scanf("%d%d",&p,&n)!=EOF)
    {
        int i;
        int x[305];
        int t;
        int tag=1;
        int temp;
        int w[304];//標記數組
        memset(w,0,sizeof(w));
        for(i=0;i<n;i++)
            scanf("%d",&x[i]);
        for(i=0;i<n;i++)
        {
            t=x[i]%p;
            if(w[t]==0)
                w[t]=1;//進行標記
            else
            {
                tag=0;
                temp=i;
                break;
            }


        }
        if(tag==0)
            printf("%d\n",temp+1);
        else
            printf("-1\n");
    }
    return 0;
}