#include <bits/stdc++.h>
using namespace std;
int x[500];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int i = -1;
  for (int j = 1; j <= m; j++) {
    int u;
    scanf("%d", &u);
    if (x[u % n] == 0)
      x[u % n]++;
    else {
      if (i == -1) i = j;
    }
  }
  printf("%d\n", i);
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int x[500];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int i = -1;
  for (int j = 1; j <= m; j++) {
    int u;
    scanf("%d", &u);
    if (x[u % n] == 0)
      x[u % n]++;
    else {
      if (i == -1) i = j;
    }
  }
  printf("%d\n", i);
}

```