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;
template <class T>
int getbit(int i, T X) {
  return (X & (1 << (i - 1)));
}
template <class T>
T onbit(int i, T X) {
  return (X | (1 << (i - 1)));
}
template <class T>
T offbit(int i, T X) {
  return (X | (1 << (i - 1)) - (1 << (i - 1)));
}
template <class T>
T sqr(T x) {
  return (x * x);
}
template <class T>
T cube(T x) {
  return (x * x * x);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int csx[4] = {0, 0, -1, 1};
int csy[4] = {-1, 1, 0, 0};
int main() {
  int a, p, n, dd[305];
  ;
  memset(dd, 0, sizeof(dd));
  cin >> p >> n;
  for (typeof(n) i = 1; i <= n; i++) {
    cin >> a;
    int pp = a % p;
    if (dd[pp]) {
      cout << i << endl;
      return 0;
    }
    dd[pp] = 1;
  }
  cout << "-1";
  return 0;
}
