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: Runtime Error)

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <numeric>
#include <bitset>

using namespace std;
using ll = long long;
using ld = long double;
const ll mod =  998244353;
const ll inf = 1e18;

#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP_FROM(i, j, n) for (ll i = (j); i < (n); ++i)
#define all(x) (x).begin(),(x).end()

template<typename T> void print_vec(vector<T> a) {int n = a.size();REP(i, n) {cout >> a[i]; if(i!=n-1) cout << " "; else cout << endl;}}
ll power(ll base, ll exponent){ if(exponent % 2){ return power(base, exponent - 1) * base % mod; }else if(exponent){ ll root_ans = power(base, exponent / 2); return root_ans * root_ans % mod; }else{ return 1; }}
ll inverse(ll x){ return power(x, mod - 2);}
ll gcd(ll a, ll b) {if(a < b) gcd(b, a);ll r;while(r=a%b) {a=b;b=r;}return b;}
template<typename T> ll sum(T begin, T end) {return accumulate(begin, end, 0ll);}
struct combination{ vector<ll> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } ll C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

int bsearch(const vector<ll>& a, ll x){
  int lft = 0;
  int rgt = a.size();
  while (rgt - lft > 1) {
    int mid = (lft + rgt) / 2;
    if(a[mid] < x) {
      lft = mid;
    }
    else {
      rgt = mid;
    }
  }
  return lft;
}

signed main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<ll> a(n);
  REP(i, n) cin >> a[i];
  ll s = sum(all(a));
  
  ll ans = 1;
  REP_FROM(i, 2, 1e6+1) {
    if(s % i != 0) continue; 
    int t = i / 2 + (i % 2 == 0? 0: 1);
    vector<ll> tmp = a;
    
    REP(j, n) {
      tmp[j] = tmp[j] % i;
    }
    sort(all(tmp));
    // REP(k, n) cout << tmp[k] << " "; cout << endl;
    int cnt = 0;
    int lft = 0;
    int rgt = n - 1;
    while(lft < rgt) {
      cnt += tmp[lft];
      tmp[rgt] += tmp[lft];
      tmp[rgt] %= i;
      tmp[lft] = 0;
      while(tmp[rgt] < 1 && rgt >= 0) {
        tmp[rgt-1] -= tmp[rgt];
        tmp[rgt-1] %= i;
        rgt--;
      }
      lft++;
      // REP(k, n) cout << tmp[k] << " "; cout << endl;
    }
    // cout << cnt << endl;
    // cout << "======" << endl;
    if(cnt <= k) ans = i;
  }
  cout << ans << endl;
  return 0; 
}