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;
struct player {
  string name;
  int score;
  bool put;
};
struct point {
  int x;
  int y;
};
int max(int arr[], int x) {
  int i = 0;
  int max = 0;
  for (int j = 1; j < x; j++) {
    if (arr[j] > arr[max]) max = j;
  }
  return max + 1;
}
int maxf(int arr[], int x) {
  int i = 0;
  int max = 0;
  for (int j = 1; j < x; j++) {
    if (arr[j] > arr[max]) max = j;
  }
  return arr[max];
}
int main() {
  int p, n, k;
  int x = 0;
  cin >> p >> n;
  int *arr = new int[p];
  for (int i = 0; i < p; i++) {
    arr[i] = 1000;
  }
  for (int i = 0; i < n; i++) {
    cin >> k;
    int r = k % p;
    if (arr[r] == 1000) {
      arr[r] = k;
      x++;
    } else {
      cout << ++x;
      return 0;
    }
  }
  cout << -1;
  return 0;
}
