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
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]

try:
    # sys.setrecursionlimit(int(pow(10,6)))
    sys.stdin = open("input.txt", "r")
    # sys.stdout = open("../output.txt", "w")
except:
    pass


n=L()[0]
A=L()
if n==1:
    print(-1)
elif n==2:
    d=max(A)-min(A)
    if d%2==0:

        # print(3)
        x=min(A)
        a=list(set([x-d,x+d//2,x+2*d]))
        print(len(a))
        print(*a)
    else:
        print(2)
        x=min(A)
        print(x-d,x+2*d)
else:
    A.sort()
    diff=[]
    for i in range(1,n):
        diff.append(A[i]-A[i-1])
    if diff.count(diff[0])==n-1:
        a=list(set([A[0]-diff[0],A[-1]+diff[0]]))
        print(len(a))
        print(*a)
    else:
        x=set(diff)
        if len(x)>2:
            print(0)
            exit()
        a=min(x)
        b=max(x)
        if a*2!=b or diff.count(a)!=n-2:
            print(0)
            exit()
        print(1)
        for i in range(n-1):
            if A[i+1]!=A[i]+a:
                print(A[i]+a)
                exit()



endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
