Problem: Little Timofey likes integers a lot. Unfortunately, he is very young and can't work with very big integers, so he does all the operations modulo his favorite prime m. Also, Timofey likes to look for arithmetical progressions everywhere.

One of his birthday presents was a sequence of distinct integers a1, a2, ..., an. Timofey wants to know whether he can rearrange the elements of the sequence so that is will be an arithmetical progression modulo m, or not.

Arithmetical progression modulo m of length n with first element x and difference d is sequence of integers x, x + d, x + 2d, ..., x + (n - 1)·d, each taken modulo m.

Input

The first line contains two integers m and n (2 ≤ m ≤ 109 + 7, 1 ≤ n ≤ 105, m is prime) — Timofey's favorite prime module and the length of the sequence.

The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai < m) — the elements of the sequence.

Output

Print -1 if it is not possible to rearrange the elements of the sequence so that is will be an arithmetical progression modulo m.

Otherwise, print two integers — the first element of the obtained progression x (0 ≤ x < m) and its difference d (0 ≤ d < m).

If there are multiple answers, print any of them.

Examples

Input

17 5
0 2 4 13 15


Output

13 2


Input

17 5
0 2 4 13 14


Output

-1


Input

5 3
1 2 3


Output

3 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct edge {
  long long to, cost;
  edge() {}
  edge(long long a, long long b) { to = a, cost = b; }
};
const long long dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
const long long mod = 1000000007;
struct mint {
  long long x = 0;
  mint(long long y = 0) {
    x = y;
    if (x < 0 || x >= mod) x = (x % mod + mod) % mod;
  }
  mint(const mint& ope) { x = ope.x; }
  mint operator-() { return mint(-x); }
  mint operator+(const mint& ope) { return mint(x) += ope; }
  mint operator-(const mint& ope) { return mint(x) -= ope; }
  mint operator*(const mint& ope) { return mint(x) *= ope; }
  mint operator/(const mint& ope) { return mint(x) /= ope; }
  mint& operator+=(const mint& ope) {
    x += ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint& ope) {
    x += mod - ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint& ope) {
    x *= ope.x, x %= mod;
    return *this;
  }
  mint& operator/=(const mint& ope) {
    long long n = mod - 2;
    mint mul = ope;
    while (n) {
      if (n & 1) *this *= mul;
      mul *= mul;
      n >>= 1;
    }
    return *this;
  }
  mint inverse() { return mint(1) / *this; }
  bool operator==(const mint& ope) { return x == ope.x; }
  bool operator!=(const mint& ope) { return x != ope.x; }
};
mint modpow(mint a, long long n) {
  if (n == 0) return mint(1);
  if (n % 2)
    return a * modpow(a, n - 1);
  else
    return modpow(a * a, n / 2);
}
istream& operator>>(istream& is, mint& ope) {
  long long t;
  is >> t, ope.x = t;
  return is;
}
ostream& operator<<(ostream& os, mint& ope) { return os << ope.x; }
ostream& operator<<(ostream& os, const mint& ope) { return os << ope.x; }
bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
void mark() { cout << "*" << endl; }
void yes() { cout << "Yes" << endl; }
void no() { cout << "No" << endl; }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long digitnum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret++;
  return ret;
}
long long digitsum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret += x % b;
  return ret;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, deque<T>& deq) {
  for (int i = 0; i < deq.size(); i++) {
    os << deq[i] << (i + 1 == deq.size() ? "" : " ");
  }
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, map<T, U>& map_var) {
  for (typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end();
       itr++) {
    os << "(" << itr->first << ", " << itr->second << ")";
    itr++;
    if (itr != map_var.end()) os << ",";
    itr--;
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& set_var) {
  for (typename set<T>::iterator itr = set_var.begin(); itr != set_var.end();
       itr++) {
    os << *itr;
    ++itr;
    if (itr != set_var.end()) os << " ";
    itr--;
  }
  return os;
}
template <typename T>
void outa(T a[], long long s, long long t) {
  for (long long i = s; i <= t; i++) {
    cout << a[i];
    if (i < t) cout << " ";
  }
  cout << endl;
}
void dump_func() { cout << endl; }
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail) {
  cout << head;
  if (sizeof...(Tail) > 0) cout << " ";
  dump_func(std::move(tail)...);
}
long long modpow(long long a, long long n, long long mod) {
  if (n == 0) return 1;
  if (n % 2) {
    return ((a % mod) * (modpow(a, n - 1, mod) % mod)) % mod;
  } else {
    return modpow((a * a) % mod, n / 2, mod) % mod;
  }
}
long long n, m;
long long a[100005];
vector<long long> vec;
pair<long long, long long> calc() {
  long long n = ((long long)(vec).size());
  if (n == 1) return pair<long long, long long>(vec[0], 1);
  long long a = vec[0], b = vec[1], x = (a - b + m) % m;
  set<long long> S;
  for (auto c : vec) S.insert(c);
  long long cnt = 0;
  for (auto c : vec)
    if (S.count((c + x) % m)) cnt++;
  long long k = n - cnt, d = modpow(k, m - 2, m) * x % m;
  while (S.count((a + m - d) % m)) a = (a + m - d) % m;
  for (long long(i) = (0); (i) <= (n - 1); (i)++)
    if (S.count((a + i * d) % m) == 0)
      return pair<long long, long long>(-1, -1);
  return pair<long long, long long>(a, d);
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> m >> n;
  for (long long(i) = (1); (i) <= (n); (i)++) cin >> a[i];
  if (n == m) {
    dump_func(0, 1);
    return 0;
  }
  if (2 * n <= m) {
    for (long long(i) = (1); (i) <= (n); (i)++) vec.push_back(a[i]);
    pair<long long, long long> ret = calc();
    if (ret.first == -1)
      dump_func(-1);
    else
      dump_func(ret.first, ret.second);
  } else {
    set<long long> S;
    for (long long(i) = (1); (i) <= (n); (i)++) S.insert(a[i]);
    for (long long(i) = (0); (i) <= (m - 1); (i)++)
      if (S.count(i) == 0) vec.push_back(i);
    pair<long long, long long> ret = calc();
    if (ret.first == -1)
      dump_func(-1);
    else
      dump_func((ret.first + ret.second * (m - n)) % m, ret.second);
  }
  return 0;
}