#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T& x) {
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
}
const int N = 310;
int a[N];
int book[N];
int n, p;
int h(int x) { return x % p; }
int main() {
  read(p);
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    int pos = h(a[i]);
    if (book[pos] == 1) {
      printf("%d\n", i);
      return 0;
    } else {
      book[pos] = 1;
    }
  }
  puts("-1");
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
template <class T>
inline void read(T& x) {
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
}
const int N = 310;
int a[N];
int book[N];
int n, p;
int h(int x) { return x % p; }
int main() {
  read(p);
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    int pos = h(a[i]);
    if (book[pos] == 1) {
      printf("%d\n", i);
      return 0;
    } else {
      book[pos] = 1;
    }
  }
  puts("-1");
  return 0;
}

```