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 arri[] = {0, 0, 1, 1, 1, -1, -1, -1}, arrj[] = {-1, 1, 0, 1, -1, 0, 1, -1};
int n, p;
int GCD(int a, int b) {
  if (!b) return a;
  return GCD(b, a % b);
}
int LCM(int a, int b) { return ((a * b) / GCD(a, b)); }
int h(int x) { return (x % p); }
int main() {
  cin >> p >> n;
  vector<bool> vis(1000);
  bool ok = true;
  int res;
  for (int i = 0; i < n && ok; i++) {
    int x;
    cin >> x;
    if (!vis[h(x)]) {
      vis[h(x)] = true;
    } else {
      res = i + 1;
      ok = false;
      break;
    }
  }
  cout << (ok ? -1 : res) << endl;
  return 0;
}
