#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int p, n;
  cin >> p >> n;
  int pos = -1;
  vector<int> v;
  int cn = 0;
  for (int i = (0); i < (n); ++i) {
    int x;
    cin >> x;
    int cnt = 0;
    for (int i1 = 0; i1 < v.size(); i1++) {
      if (v[i1] == x % p && cn == 0) {
        cout << i + 1;
        cnt++;
        break;
      }
    }
    if (cnt == 0) {
      v.push_back(x % p);
    } else {
      if (cn == 0) {
        cn++;
        pos = 0;
      }
    }
  }
  if (pos == -1) {
    cout << -1;
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int p, n;
  cin >> p >> n;
  int pos = -1;
  vector<int> v;
  int cn = 0;
  for (int i = (0); i < (n); ++i) {
    int x;
    cin >> x;
    int cnt = 0;
    for (int i1 = 0; i1 < v.size(); i1++) {
      if (v[i1] == x % p && cn == 0) {
        cout << i + 1;
        cnt++;
        break;
      }
    }
    if (cnt == 0) {
      v.push_back(x % p);
    } else {
      if (cn == 0) {
        cn++;
        pos = 0;
      }
    }
  }
  if (pos == -1) {
    cout << -1;
  }
}

```