Everybody knows what an arithmetic progression is. Let us remind you just in case that an arithmetic progression is such sequence of numbers a1, a2, ..., an of length n, that the following condition fulfills: 

a2 - a1 = a3 - a2 = a4 - a3 = ... = ai + 1 - ai = ... = an - an - 1.

For example, sequences [1, 5], [10], [5, 4, 3] are arithmetic progressions and sequences [1, 3, 2], [1, 2, 4] are not.

Alexander has n cards containing integers. Arthur wants to give Alexander exactly one more card with a number so that he could use the resulting n + 1 cards to make an arithmetic progression (Alexander has to use all of his cards).

Arthur has already bought a card but he hasn't written a number on it. Help him, print all integers that you can write on a card so that the described condition fulfilled.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of cards. The next line contains the sequence of integers — the numbers on Alexander's cards. The numbers are positive integers, each of them doesn't exceed 108.

Output

If Arthur can write infinitely many distinct integers on the card, print on a single line -1.

Otherwise, print on the first line the number of integers that suit you. In the second line, print the numbers in the increasing order. Note that the numbers in the answer can exceed 108 or even be negative (see test samples).

Examples

Input

3
4 1 7


Output

2
-2 10


Input

1
10


Output

-1


Input

4
1 3 5 9


Output

1
7


Input

4
4 3 4 5


Output

0


Input

2
2 4


Output

3
0 3 6
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O2")
const long long MOD = 1e9 + 7;
const double PI = acos(-1);
long long add(long long x, long long y) {
  long long res = x + y;
  return (res >= MOD ? res - MOD : res);
}
long long mul(long long x, long long y) {
  long long res = x * y;
  return (res >= MOD ? res % MOD : res);
}
long long sub(long long x, long long y) {
  long long res = x - y;
  return (res < 0 ? res + MOD : res);
}
long long power(long long a, long long b, long long m = MOD) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % m;
    b >>= 1;
    a = (1ll * a * a) % m;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isprime(long long n) {
  if (n < 2) return 0;
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) return 0;
    i++;
  }
  return 1;
}
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
double distform(long long x, long long y, long long z, long long w) {
  return sqrt(1. * pow(x - z, 2) + 1. * pow(y - w, 2));
}
long long dx[] = {-1, 1, 0, 0, -1, 1, 1, -1};
long long dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
const long long MAXN = 2e5 + 10;
const long long inf = 1e18;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long intlog(double base, double x) {
  return (long long)(log(x) / log(base));
}
bool isvalid(long long x, long long y, long long n, long long m) {
  return x >= 0 and x < n and y >= 0 and y < m;
}
template <typename T>
void read(T &x) {
  for (long long i = 0; i < ((long long)(x).size()); ++i) cin >> x[i];
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  read(a);
  sort(a.begin(), a.end());
  unordered_map<long long, long long> diff;
  for (long long i = 1; i < n; ++i) {
    diff[a[i] - a[i - 1]]++;
  }
  if (((long long)(diff).size()) > 2) {
    cout << 0 << endl;
    return;
  }
  long long minn = inf;
  long long diffatmin;
  long long other = 1;
  for (auto &x : diff) {
    if (minn > x.second) {
      minn = x.second;
      diffatmin = x.first;
    }
    if (minn == x.second) {
      if (diffatmin < x.first) diffatmin = x.first;
    }
  }
  for (auto &x : diff) {
    if (x.first == diffatmin) continue;
    other = x.first;
  }
  if (((long long)(diff).size()) > 1 and minn > 1) {
    cout << 0 << endl;
    return;
  }
  {
    cerr << " ["
         << "(diffatmin % 2 != 0)"
         << ": " << (diffatmin % 2 != 0) << "]\n";
  }
  {
    cerr << " ["
         << "(diffatmin / 2 != other)"
         << ": " << (diffatmin / 2 != other) << "]\n";
  }
  {
    cerr << " ["
         << "(other != 0)"
         << ": " << (other != 0) << "]\n";
  }
  if (((long long)(diff).size()) > 1 and
      (diffatmin % 2 != 0 or diffatmin / 2 != other or other == 0)) {
    cout << 0 << endl;
    return;
  }
  if (((long long)(diff).size()) == 2) {
    cout << 1 << endl;
    for (long long i = 1; i < n; ++i) {
      if (a[i] - a[i - 1] == diffatmin) {
        cout << a[i - 1] + diffatmin / 2 << endl;
        return;
      }
    }
  }
  if (a.size() == 1) {
    cout << -1 << endl;
    return;
  }
  bool ok = false;
  bool x = ((a[1] - a[0]) % 2 == 0);
  ok = ((long long)(a).size()) == 2 and ((a[1] - a[0]) % 2 == 0);
  if (diff.begin()->first == 0) {
    cout << 1 << endl;
    cout << a[0] << endl;
    return;
  }
  cout << 2 + ok << endl;
  cout << a[0] - diff.begin()->first << " ";
  if (ok) {
    cout << a[0] + (a[1] - a[0]) / 2 << " ";
  }
  cout << a[n - 1] + diff.begin()->first << " ";
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}
