#include <bits/stdc++.h>
int i, j;
using namespace std;
int main() {
  vector<long long> v(100000, 0);
  vector<long long> ar;
  int p, n, m = 11111110;
  cin >> p >> n;
  for (i = 0; i < n; i++) cin >> v[i];
  for (i = 0; i < n; i++) {
    long long x;
    x = v[i] % p;
    ar.push_back(x);
  }
  int c = 0, t = 1, k;
  for (i = 0; i < n - 1; i++) {
    for (j = i + 1; j < n; j++) {
      if (ar[i] == ar[j]) {
        k = j + 1;
        m = min(k, m);
        c++;
        t = 0;
      }
    }
  }
  if (c == 0)
    cout << "-1\n";
  else
    cout << m << endl;
}

### 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 i, j;
using namespace std;
int main() {
  vector<long long> v(100000, 0);
  vector<long long> ar;
  int p, n, m = 11111110;
  cin >> p >> n;
  for (i = 0; i < n; i++) cin >> v[i];
  for (i = 0; i < n; i++) {
    long long x;
    x = v[i] % p;
    ar.push_back(x);
  }
  int c = 0, t = 1, k;
  for (i = 0; i < n - 1; i++) {
    for (j = i + 1; j < n; j++) {
      if (ar[i] == ar[j]) {
        k = j + 1;
        m = min(k, m);
        c++;
        t = 0;
      }
    }
  }
  if (c == 0)
    cout << "-1\n";
  else
    cout << m << endl;
}

```