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: Accepted)

#include<bits/stdc++.h>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=m-1; i>=n; i--)
#define SORT(v,n) do{sort(v,v+n);reverse(v,v+n);}while(0)
#define all(x) (x).begin(),(x).end()
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
typedef pair<ll, ll> LP;

ll POW(ll x,ll n){
  x%=MOD;
  if(n==0)return 1;
  if(n%2==0)return POW(x*x,n/2)%MOD;
  return x%MOD*POW(x,n-1)%MOD;
}

ll POW2(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW2(x*x,n/2);
  return x*POW2(x,n-1);
}

ll POW3(ll x,ll n,ll m){
  x%=m;
  if(n==0)return 1;
  if(n%2==0)return POW3(x*x,n/2,m)%m;
  return x*POW3(x,n-1,m)%m;
}

ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}

ll lcm(ll u, ll v) {
  return u*v/gcd(u,v);
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll KAI2(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI2(m-1);
}

ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI(m)%MOD*modinv(KAI(n)%MOD*KAI(m-n)%MOD)%MOD;
}

ll COM2(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI2(m)/KAI2(n)/KAI2(m-n);
}

ll DEC(ll x,ll m,ll n)//xのm進数でのx^nの位の値
{
  return x%POW2(m,n+1)/POW2(m,n);
}

ll keta(ll x,ll n)//xのn進数での桁数
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}

ll DIV(ll x,ll n)//x!のnで割り切れる回数
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}

ll ORD(ll x,ll n)//xのnで割り切れる回数
{
  if(x==0)return INF;
  if(x%n!=0)return 0;
  return 1+ORD(x/n,n);
}

ll SGS(ll x,ll y, ll m)//1+x+…+x^(y-1)をmで割った余り
{
  if(y==0)return 0;
  if(y%2==0){
    return (1+POW3(x,y/2,m))*SGS(x,y/2,m)%m;
  }
  return (1+x*SGS(x,y-1,m))%m;
}

ll SSGS(ll x,ll y,ll m)//Σ[k=1→y](1+x+…+x^(k-1))をmで割った余り
{
  if(y==0)return 0;
  if(y==1)return 1;
  if(y%2==0){
    return (SSGS(x,y/2,m)*(POW3(x,y/2,m)+1)%m+SGS(x,y/2,m)*y/2%m)%m;
  }
  return (SSGS(x,y-1,m)*x%m+y)%m;
}

struct UnionFind
{
  vector<int> par;
  vector<int> sizes;
  UnionFind(int n) : par(n), sizes(n, 1) {
    rep(i,n) par[i] = i;
  }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (sizes[x] < sizes[y]) swap(x, y);
    par[y] = x;
    sizes[x] += sizes[y];
  }
  bool same(int x, int y) {
    return find(x) == find(y);
  }
  int size(int x) {
    return sizes[find(x)];
  }
};

int main()
{
  ll n,k,a[510],m=0,q,c=0,ans=1,b[510],t=0,l=0,o,u=0;
  double x;
  cin >> n >> k;
  rep(i,n)cin >> a[i];
  rep(i,n)m+=a[i];
  x=m;
  x=sqrt(x);
  q=x;
  FOR(i,1,q+2){
    if(m%i==0){
      rep(j,n)b[j]=a[j]%i;
      sort(b,b+n);
      t=k;
      c=0;
      rep(j,n){
        c+=b[j];
        if(c>k){
          l=c-b[j];
          o=j-1;
          goto A;
        }
      }
      l=c;
      o=n-1;
      A:;
      u=0;
      FOR(j,o+1,n){
        u+=i-b[j];
      }
      if(u<=c){
        ans=max(ans,i);
      }
      c=0;
      
      
      rep(j,n)b[j]=a[j]%(m/i);
      sort(b,b+n);
      t=k;
      c=0;
      rep(j,n){
        c+=b[j];
        if(c>k){
          l=c-b[j];
          o=j-1;
          goto B;
        }
      }
      l=c;
      o=n-1;
      B:;
      u=0;
      FOR(j,o+1,n){
        u+=(m/i)-b[j];
      }
      if(u<=c){
        ans=max(ans,(m/i));
      }
      c=0;
    }
  }
  printf("%lld",ans);
}