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;
int main() {
  long long int a, b, c = 0, d = 0, m, i, n, j, p, t = 0, k, array[100000 + 10],
                      count[100000], q, sum1 = 0, cnt = 0, sum = 0, flag, ans,
                      f = 0, l;
  long long int st, ct, qt;
  string str, ptr, sp[5000];
  set<pair<long long int, long long int>> vecs;
  vector<long long int> vi(100000, 0);
  stack<long long int> stck;
  vector<char> vc;
  memset(count, 0, sizeof(count));
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> array[i];
  }
  for (i = 1; i <= m; i++) {
    if (count[array[i] % n] == 0) {
      l = array[i] % n;
      count[l]++;
    } else {
      ans = i;
      cout << ans << endl;
      return 0;
    }
  }
  cout << -1 << endl;
}
