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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline long long two(long long n) { return 1 << n; }
inline long long test(long long n, long long b) { return (n >> b) & 1; }
inline void setBit(long long &n, long long b) { n |= two(b); }
inline void unsetBit(long long &n, long long b) { n &= ~two(b); }
inline long long last_bit(long long n) { return n & (-n); }
inline long long ones(long long n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
inline bool sortDown(long long x, long long y) { return x > y; }
template <class T>
void chmax(T &a, const T &b) {
  a = max(a, b);
}
template <class T>
void chmin(T &a, const T &b) {
  a = min(a, b);
}
bool wayToSort(long long i, long long j) { return i > j; }
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
bool sortbysecdesc(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
  return a.second > b.second;
}
long long a[100010];
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (n); i++) cin >> a[i];
  if (n == 1) {
    cout << -1 << endl;
    return 0;
  }
  sort(a, a + n);
  bool equal = true;
  for (long long i = (0); i <= (n - 2); i++) {
    if (a[i] != a[i + 1]) {
      equal = false;
      break;
    }
  }
  if (equal) {
    cout << 1 << endl << a[0] << endl;
    return 0;
  }
  if (n == 2) {
    if ((a[1] - a[0]) % 2 == 0) {
      cout << 3 << endl
           << 2 * a[0] - a[1] << " " << a[0] + (a[1] - a[0]) / 2 << " "
           << 2 * a[1] - a[0] << endl;
    } else {
      cout << 2 << endl << 2 * a[0] - a[1] << " " << 2 * a[1] - a[0] << endl;
    }
    return 0;
  }
  set<long long> diff;
  for (long long i = (1); i <= (n - 1); i++) diff.insert(a[i] - a[i - 1]);
  if (diff.size() == 1) {
    cout << 2 << endl
         << a[0] - (a[1] - a[0]) << " " << a[n - 1] + (a[1] - a[0]) << endl;
    return 0;
  } else if (diff.size() == 2) {
    long long d[3];
    long long i = 0;
    for (auto u : diff) d[i++] = u;
    if (d[1] != 2 * d[0]) {
      cout << 0 << endl;
      return 0;
    } else {
      std::vector<long long> v;
      long long count = 0;
      for (long long i = (1); i <= (n - 1); i++) {
        if (a[i] - a[i - 1] == d[1]) {
          if (count < 1) {
            v.push_back(a[i - 1] + d[0]);
            count++;
          } else {
            cout << 0 << endl;
            return 0;
          }
        }
      }
      cout << 1 << endl << v[0] << endl;
      return 0;
    }
  }
  cout << 0 << endl;
  return 0;
}
