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)

#pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define SIZE 300005
#define MOD 1000000007LL
#define EPS 1e-10
#define INF 2147483647
#define LLINF 9223372036854775807LL
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define DOWN(i,b,a) for(int i=b;i>=a;i--)
#define SET(a,c) memset(a,c,sizeof a)
#define BIT(i,j) ((i)>>(j))&1
#define ALL(o) (o).begin(), (o).end()
#define ERASE(o) (o).erase(unique((o).begin(),(o).end()), (o).end())
#define SQ(x) ((x)*(x))
using namespace std;
typedef int_fast64_t ll;
typedef pair<ll,ll> Pll;
typedef pair<int, int> Pii;
typedef pair<double, double> Pdd;
typedef complex<double> dcomplex;
template<typename T> inline void priv(vector<T>a){REP(i,a.size()){cerr<<a[i]<<((i==a.size()-1)?"\n":" ");}}
inline void fastio(){cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15);}
ll gcd(ll a,ll b){ll c=max(a,b);ll d=min(a,b);return c==0||d==0?c:gcd(c%d,d);}
ll lcm(ll a,ll b){return a==0||b==0?0:a*b/gcd(a,b);}
ll modfact(ll a){ll b=1;FOR(i,2,a)b=b*i%MOD;return b;}
ll modpow(ll a, ll n){ll b=1;while (n>0){if(n&1)b=b*a%MOD;a=a*a%MOD;n>>=1;}return b;}
ll modcomb(ll n, ll k){ll b=1;k=min(n-k,k);DOWN(i,n,n-k+1)b=b*i%MOD;return b*modpow(modfact(k),MOD-2)%MOD;}

size_t i, j;
ll N, K, S = 0;
ll A[500];

bool solve(ll m) {
    vector<ll> B(N, 0);
    for(i = 0; i < N; i++) B[i] = A[i] % m;
    sort(ALL(B));
    j = N - 1;
    ll cnt = 0, buff = 0;
    for(i = 0; i <= j; i++) {
        if(B[i] % m == 0) continue;
        buff += B[i];
        cnt += B[i];
        B[i] = 0;
        while(buff && j >= 0) {
            if(B[j] % m == 0) {
                j--;
                continue;
            }
            if(m - B[j] > buff) {
                B[j] += buff;
                buff = 0;
            } else {
                buff -= (m - B[j]);
                B[j] = m;
                j--;
            }
        }
    } 
    return cnt <= K;
}

int main() {
    fastio();
    cin >> N >> K;
    REP(i, N) {
        cin >> A[i];
        S += A[i];
    }
    ll x = 1;
    vector<ll> cand;
    while(x * x <= S) {
        if(S % x == 0) {
            cand.push_back(x); 
            cand.push_back(S / x);
        }
        x++;
    }
    sort(ALL(cand)); reverse(ALL(cand));
    for(auto& c : cand) if(solve(c)) {
        cout << c << endl;
        break;
    }
	return 0;
}