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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i,n) for(int64_t i=0;i<(n);i++)
#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REV(c) reverse((c).begin(),(c).end())
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define VIN(n,v) for(int64_t i=0;i<(n);i++){cin >> (v).at(i);}

using namespace std;
typedef int64_t I;
typedef pair<I,I> P;
typedef vector<I> V;

I MOD = pow(10,9)+7;

// 最大公約数
I gcd(I a, I b) {
  return b != 0 ? gcd(b, a % b) : a;
}

// 最小公倍数
I lcm(I a, I b) {
  return a * b / gcd(a, b);
}

int ctoi(char c){
  return c - '0';
}

V enum_div(auto n)//nの約数を列挙
{
  V ret;
  for(int i=1 ; i*i<=n ; ++i)
  {
    if(n%i == 0)
    {
      ret.push_back(i);
      if(i!=1 && i*i!=n)
      {
        ret.push_back(n/i);
      }
    }
  }
  return ret;
}

int main(){
  I n,k;
  cin >> n >> k;
  V a(n);
  VIN(n,a);
  I t=0;
  REP(i,n){
    t += a.at(i);
  }
  V div = enum_div(t);
  div.push_back(t);
  I an=0;
  REP(i,div.size()){
    I diff=0;
    V b(n);
    b = a;
    REP(j,a.size()){
      b.at(j) = a.at(j) % div.at(i);
    }
    SORT(b);
    REP(j,b.size()){
      I base = b.at(j) % div.at(i);
      diff += base;
      I c = b.size() - 1;
      while (true){
        if (base > 0){
          if (b.at(c) != 0){
            I chk = min(div.at(i) - b.at(c),base);
            b.at(c) += chk;
            base -= chk;
            b.at(c) %= div.at(i);
            c--;
          }else{
            break;
          }
        }else{
          break;
        }
      }
    }
    if (diff <= k){
      an = max(an,div.at(i));
    }
  }
  cout << an << endl;
  return 0;
}