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 const N = 1e6 + 3;
long long MOD = 1e9 + 7, fact[N];
const long long INF = (long long)1e18;
const long double PI = 3.14159265358979;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int p, n;
  cin >> p >> n;
  set<int> s;
  int ans = 0;
  int tmp = 0, flag = 0;
  while (n--) {
    int x;
    tmp++;
    cin >> x;
    if (s.find(x % p) != s.end() && flag == 0) {
      ans = tmp;
      flag = 1;
    } else {
      s.insert(x % p);
    }
  }
  if (ans == 0) {
    cout << -1 << '\n';
  } else {
    cout << ans << '\n';
  }
  return 0;
}
