#include <bits/stdc++.h>
int main() {
  int vis[1000];
  memset(vis, 0, sizeof(vis));
  int p, n, i, x, flag = -1;
  scanf("%d %d", &p, &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    vis[x % p]++;
    if (vis[x % p] > 1 && flag == -1) flag = i;
  }
  printf("%d\n", flag);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

Examples

Input

10 5
0
21
53
41
53


Output

4


Input

5 5
0
1
2
3
4


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int vis[1000];
  memset(vis, 0, sizeof(vis));
  int p, n, i, x, flag = -1;
  scanf("%d %d", &p, &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    vis[x % p]++;
    if (vis[x % p] > 1 && flag == -1) flag = i;
  }
  printf("%d\n", flag);
  return 0;
}

```