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 PHP (5.6.30) (judge verdict: Time Limit Exceeded)

<?php
fscanf(STDIN, "%d %d", $n, $k);
$list = explode(" ", trim(fgets(STDIN)));
$total=array_sum($list);

//約数探し
for($i=$total; $i>=0; $i--){
  if($total%$i==0){
    //約数なら
    $modList=array();
    foreach($list as $v){
      if($v%$i!=0){
  		$modList[count($modList)]=$v%$i;
      }
	}
    sort($modList);

    $tmp=0;
    $sum=0;
    $endKey=count($modList)-1;
    for($j=0; $modList[$j]!=0; $j++){
      $sum+=$modList[$j];
      if(($modList[$endKey]+$tmp+$modList[$j])>=$i){
        $modList[$endKey]=0;
        $tmp=($modList[$endKey]+$tmp+$modList[$j])%$i;
        $endKey--;
      }else{
        $modList[$endKey]+=$modList[$j];
      }  
    }
    //sumがk（移動上限）以下なら、iを表示して、終了
    if($sum <= $k){
      echo $i;
      exit;
    }
  }
}