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>
#pragma comment(linker, "/stack:227420978")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long MOD = 1000000007LL, INF = 1e9, LINF = 1e18;
const long double PI = 3.141592653589793116, EPS = 1e-9,
                  GOLD = ((1 + sqrt(5)) / 2);
long long keymod[] = {1000000007LL, 1000000009LL, 1000000021LL, 1000000033LL};
vector<long long> HashMod(keymod, keymod + sizeof(keymod) / sizeof(long long));
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
auto TimeStart = chrono::steady_clock::now();
auto TimeEnd = chrono::steady_clock::now();
long long p, n, a;
map<long long, long long> Map;
void ControlIO(int argc, char* argv[]);
void TimerStart();
void TimerStop();
void Exit();
void Input() { cin >> p >> n; }
void Solve() {
  for (long long i = 1; i <= n; i++) {
    cin >> a;
    a %= p;
    if (Map[a]) {
      cout << i;
      return;
    }
    Map[a]++;
  }
  cout << "-1";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ControlIO(argc, argv);
  Input();
  TimerStart();
  Solve();
  TimerStop();
  return 0;
}
void ControlIO(int argc, char* argv[]) {}
void TimerStart() {}
void TimerStop() {}
void Exit() {
  TimerStop();
  exit(0);
}
