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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from bisect import bisect_left, bisect_right; from collections import Counter, defaultdict, deque, OrderedDict; from copy import deepcopy; from fractions import gcd; from functools import lru_cache, reduce; from math import ceil, floor; from sys import stdin, setrecursionlimit
import heapq as hq; import itertools as its; import operator as op


# globals
inf = float('inf')
input = stdin.readline


def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    divisors.sort()
    return divisors


def main():
    n, k = get_li()
    A = get_li()
    ds = make_divisors(sum(A))
    ans = 1
    for d in ds[1:]:
        print(d)
        pdiff = [0] * n
        ndiff = [0] * n
        for i, a in enumerate(A):
            pdiff[i] = a % d
            ndiff[i] = a % d - d
        hoge = list(sorted(zip(pdiff, ndiff)))
        ps = [h[0] for h in hoge]
        ns = [h[1] for h in hoge]
        pacc = list(its.accumulate(ps))
        nacc = list(reversed(list(its.accumulate(reversed(ns)))))
        print(pacc)
        print(nacc)
        print()
        if min([abs(p + n) for p, n in zip(pacc, nacc)]) <= k * 2:
            ans = d
    print(ans)
    return


def get_int():
    return int(input())


def get_float():
    return float(input())


def get_str():
    return input().strip()


def get_li():
    return [int(i) for i in input().split()]


def get_lf():
    return [float(f) for f in input().split()]


def get_lc():
    return list(input().strip())


def gets(n, types, sep=None):
    if len(types) == 1:
        return [types[0](input().strip()) for _ in range(n)]
    return list(zip(*(
        [t(x) for t, x in zip(types, input().strip().split(sep=sep))]
        for _ in range(n)
    )))


if __name__ == '__main__':
    setrecursionlimit(4100000)
    main()