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 int inp() {
  int n, ch;
  while ((ch = getchar()) < '0')
    ;
  n = (ch - '0');
  while ((ch = getchar()) >= '0') n = (n << 3) + (n << 1) + (ch - '0');
  return n;
}
short gcd(short a, short b) {
  while (a != 0) {
    unsigned long long int temp = a;
    a = b % a;
    b = temp;
  }
  return b;
}
int main() {
  int p = inp();
  int n = inp();
  bool check[p];
  for (int i = 0; i < p; i++) check[i] = false;
  for (int i = 0; i < n; i++) {
    int k = inp();
    if (check[k % p]) {
      printf("%d", i + 1);
      return 0;
    } else
      check[k % p] = true;
  }
  printf("-1");
  return 0;
}
