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;
inline void scanInt(int &x) {
  register int c = getchar();
  x = 0;
  while (c < 48 || c > 57) c = getchar();
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
}
inline void scanIntNeg(int &x) {
  register int c = getchar();
  x = 0;
  bool neg = false;
  while (c < 48 || c > 57) {
    if (c == '-') neg = true;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x *= -1;
}
set<int> presenti;
int main(int argc, const char *argv[]) {
  int P, N;
  scanInt(P);
  scanInt(N);
  for (int i = 0; i < N; i++) {
    int att;
    scanInt(att);
    if (presenti.find(att % P) == presenti.end())
      presenti.insert(att % P);
    else {
      cout << i + 1;
      return 0;
    }
  }
  cout << -1;
  return 0;
}
