#include <bits/stdc++.h>
const int N = 100000;
using namespace std;
bool h[N] = {0};
int main(int argc, char *argv[]) {
  int p, n, x;
  cin >> p >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (h[x % p]) {
      cout << i << '\n';
      return 0;
    } else
      h[x % p] = 1;
  }
  cout << -1 << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const int N = 100000;
using namespace std;
bool h[N] = {0};
int main(int argc, char *argv[]) {
  int p, n, x;
  cin >> p >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (h[x % p]) {
      cout << i << '\n';
      return 0;
    } else
      h[x % p] = 1;
  }
  cout << -1 << '\n';
  return 0;
}

```