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 a[10001], b[1001];
bool stug[10001];
bool parz(int k) {
  for (int j = 2; j <= sqrt(k); j++) {
    if (k % j == 0) return false;
  }
  return true;
}
struct delta {
  int x;
  int y;
};
delta tri[100001];
int main() {
  long long n, m, k = 0, l = 0, j = 0;
  char s[1001], s1[1001];
  long long ans = 0;
  int i = 0, mx = 0;
  int mxi, mxii;
  int x, y, t;
  bool o = false;
  map<int, int> mp;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> x;
    mp[x % n]++;
    if (mp[x % n] > 1) {
      o = true;
      cout << i + 1 << endl;
      return 0;
    }
  }
  if (!o) {
    cout << -1 << endl;
  }
  return 0;
}
