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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int i, n, t, j = 0, p, x, a = 0, b, l = 0, r = 0, y, k, c, sum = 0,
                         m, d, cnts = 0;
  cin >> p >> n;
  int ara[400] = {};
  l = 0;
  x = 0;
  while (n--) {
    l++;
    cin >> a;
    if (ara[a % p] == 1 && x == 0) {
      x = l;
    } else {
      ara[a % p] = 1;
    }
  }
  if (x)
    cout << x << endl;
  else
    cout << -1 << endl;
}
