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 9.2.1) (judge verdict: Wrong Answer)

#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <iomanip> // setprecision
#include <complex> // complex
#include <math.h> 
#include <climits> 
#include <assert.h>
#include <random>
#define rep(i,n) for(int i=0; i<(n); i++)
using namespace std;
using ll = long long;
using P = pair<int,int>;
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr ll mod = 1'000'000'007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[8] = {1, 0, -1, 0,1,1,-1,-1};
const int dy[8] = {0, 1, 0, -1,1,-1,1,-1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// ---------------------------------------------------------------------------

int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N,K;
  cin >> N >> K;
  vector<int> A(N);
  ll sum = 0;
  for(int i=0; i<N; i++){
    cin >> A[i];
    sum += A[i];
  }
  auto suzu = [&](int d){
    vector<int> rem(N);
    for(int i=0; i<N; i++){
      rem[i] = A[i]%d;
    }
    sort(rem.begin(),rem.end());
    vector<int> S(N+1,0);
    for(int i=0; i<N; i++){
      S[i+1] = S[i] + rem[i];
    }
    vector<int> suf(N+1,0);
    for(int i=0; i<N; i++){
      suf[i+1] = suf[i] + (d-rem[i]);
    }
    for(int i=0; i<N; i++){      
      if(max(S[i],suf[N]-suf[i]) <= K) return true;
    }
    return false;
  };
  ll ans = 0;
  for(ll i=1; i*i<=sum; i++){
    if(sum%i == 0){
      if(suzu(i)) chmax(ans,i);
      if(suzu(sum/i)) chmax(ans,sum/i);
    }
  }
  cout << ans << "\n";
  return 0;
}