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;
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int minn = 1e5 + 9;
const int N = 5e3 + 123;
const int NN = 5e5 + 123;
const int PRI = 555557;
const int pri = 997;
int tests = 1, t[NN], u[NN], p;
inline void Solve() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> p >> n;
  for (int i = 1; i <= n; i++) cin >> t[i];
  for (int i = 1; i <= n; i++) {
    if (u[t[i] % p]) {
      cout << i;
      exit(0);
    }
    u[t[i] % p] = 1;
  }
  cout << -1;
}
int main() {
  while (tests--) {
    Solve();
  }
}
