Problem:
Score : 500 points

Problem Statement

We have a sequence of N integers: A_1, A_2, \cdots, A_N.

You can perform the following operation between 0 and K times (inclusive):

Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.

Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.

Constraints

2 \leq N \leq 500

1 \leq A_i \leq 10^6

0 \leq K \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_{N-1} A_{N}

Output

Print the maximum possible positive integer that divides every element of A after the operations.

Sample Input 1

2 3
8 20

Sample Output 1

7

7 will divide every element of A if, for example, we perform the following operation:

Choose i = 2, j = 1. A becomes (7, 21).

We cannot reach the situation where 8 or greater integer divides every element of A.

Sample Input 2

2 10
3 5

Sample Output 2

8

Consider performing the following five operations:

Choose i = 2, j = 1. A becomes (2, 6).

Choose i = 2, j = 1. A becomes (1, 7).

Choose i = 2, j = 1. A becomes (0, 8).

Choose i = 2, j = 1. A becomes (-1, 9).

Choose i = 1, j = 2. A becomes (0, 8).

Then, 0 = 8 \times 0 and 8 = 8 \times 1, so 8 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

Sample Input 3

4 5
10 1 2 22

Sample Output 3

7

Sample Input 4

8 7
1 7 5 6 8 2 6 5

Sample Output 4

5

Solution in Python (3.4.3) (judge verdict: Accepted)

N, K = map(int, input().split())
A = list(map(int, input().split()))
# A.sort()
totalA = sum(A)
# print (totalA)

def make_divisors(n): #nの約数をO(root(n))で全列挙、sortして取り出すときはO(root(n) log(n))
    import math
    divisors = []
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0: #割り切れるとき
            divisors.append(i)
            if i != n//i:
                divisors.append(n//i)
    divisors.sort(reverse = True)
    return divisors

divisors_lst = make_divisors(totalA)
# print (divisors_lst)
for i in divisors_lst: #答えの候補を全探索
    lst = []
    for j in A:
        lst.append(j % i)
    lst.sort()
    total_lst = 0
    for j in lst:
        total_lst += i - j
    tmp = 0
    for k in lst:
        tmp += k
        total_lst -= (i - k)
        if max(tmp, total_lst) <= K:
            print (i)
            exit()