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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */


int n;
int a[500];
int ma[500];

int getNeed(int t)
{
    uint64_t msum;
    for(int i=0;i<n;i++)
    {
        ma[i] = a[i]%t;
    }
    qsort(ma, n,sizeof(int), acs);
    msum = 0;
    for(int i=0;i<n;i++)
    {
        //printf("%d%c",ma[i],i==n-1?'\n':' ');
        msum += ma[i];
    }
    uint64_t need;
    need = 0;
    for(int i=0;i<n-(msum/t);i++)
    {
        need += ma[i];
    }
    return need;
}


int main(void)
{
    int k;
    uint64_t sum;
    scanf("%d",&n);
    scanf("%d",&k);

    sum = 0;
    for(int i=0;i<n;i++)
    {
        scanf("%d", &(a[i]));
        sum += a[i];
    }

    uint64_t ans;
    ans = 1;
    for(int i=1; sum >= i*i;i++)
    {
        if( sum %i == 0)
        {
            int n;
            n = getNeed(i);
            if( n<=k) ans = ans < i ? i: ans;
            n = getNeed(sum/i);
            if( n<=k) ans = ans < sum/i ? sum/i: ans;
        }
    }

    printf("%ld\n",ans);
    
    return 0;
}