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;
set<char> st;
set<char>::iterator it;
map<long long, long long> mp;
map<long long, long long>::iterator itr;
vector<string> vrat;
vector<string> vwom_chi;
vector<string> vmen;
int arr[3][3], a[3][3];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  long long a, b, l, arr[1000], ans[1000] = {0}, s = -1;
  cin >> a >> b;
  for (int i = 0; i < b; i++) cin >> arr[i];
  for (int i = 0; i < b; i++) {
    if (ans[(arr[i] % a)] > 0) {
      cout << i + 1 << endl;
      s = 0;
      break;
    }
    ans[(arr[i] % a)]++;
  }
  if (s == -1) cout << "-1" << endl;
}
