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
#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 0, 1, 1, 1, 0, -1, -1}, dy[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int cx[] = {1, 1, 2, 2, -1, -1, -2, -2}, cy[] = {2, -2, 1, -1, 2, -2, 1, -1};
template <class T>
int len(const T& c) {
  return (int)c.size();
}
long long stoi(string s) {
  long long n = 0;
  for (__typeof(len(s)) i = 0; i < (len(s)); i++)
    n = (n << 3) + (n << 1) + (s[i] - 48);
  return n;
}
string itos(long long n) {
  if (n == 0) return "0";
  string s;
  while (n) {
    s += (n % 10 + 48);
    n /= 10;
  }
  reverse((s).begin(), (s).end());
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  long p, n;
  while (cin >> p >> n) {
    set<long> string;
    for (__typeof(n) i = 1; i <= (n); i++) {
      long a, b;
      b = string.size();
      cin >> a;
      string.insert(a % p);
      if (b == string.size()) {
        cout << i << endl;
        return 0;
      }
    }
    cout << -1 << endl;
  }
  return 0;
}
