#include <bits/stdc++.h>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
set<int> s;
int main() {
  int i, j, p, n, prev = -1;
  cin >> p >> n;
  for (j = 0; j < n; j++) {
    int x;
    cin >> x;
    int k = x % p;
    if (prev != -1) continue;
    if (s.find(k) != s.end()) {
      prev = j + 1;
    } else
      s.insert(k);
  }
  cout << prev;
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
set<int> s;
int main() {
  int i, j, p, n, prev = -1;
  cin >> p >> n;
  for (j = 0; j < n; j++) {
    int x;
    cin >> x;
    int k = x % p;
    if (prev != -1) continue;
    if (s.find(k) != s.end()) {
      prev = j + 1;
    } else
      s.insert(k);
  }
  cout << prev;
  return 0;
}

```