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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define df 0
int dc=100000;
typedef long int li;
#define MAX 512
long int n,a[MAX],k;
li r[MAX];
li s=0;
int ok(li d);
int comp(const void* a,const void* b){
  li x=*((li*)a),y=*((li*)b);
  return y-x;
}

int main(){
  if(df)printf("*** debug mode ***\n");
  int i;
  scanf("%d%d",&n,&k);
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
    s+=a[i];
  }
  li max=0;
  for(i=1;i<=sqrt(s);i++){
    if(s%i)continue;
    if(ok(s/i)){max=s/i;break;}
    if(ok(i)) max=i;
  }
  

  printf("%ld",max);
  return 0;
}
int ok(li d){
  if(df)printf("%ld:\n",d);
  int i;
  li sum=0;
  for(i=0;i<n;i++){
    r[i]=a[i]%d;
    sum+=r[i];
  }
  if(df){
    for(i=0;i<n;i++)printf("%ld ",r[i]);
    printf(":r\n");
  }
  li x=sum/d;
  qsort(r,n,sizeof(li),comp);
  if(df){
    for(i=0;i<n;i++)printf("%ld ",r[i]);
    printf(":r (orderd)\n");
  }
  
  sum=0;
  for(i=0;i<x;i++){
    sum+=d-r[i];
  }
  if(df)printf("need: %ld\n",sum);

  return (sum<=k)?1:0;
}

/// confirm df==0 ///