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(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
#define Fast_cin ios::sync_with_stdio(false), cin.tie(0);
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
using namespace std;

typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

template <typename T>
struct hash_map_t {
    vector <T> v, val, nxt;
    vector <int> head;
    int mod, tot, lastv;
    T lastans;
    bool have_ans;

    hash_map_t (int md = 0) {
        head.clear(); v.clear(); val.clear(); nxt.clear(); tot = 0; mod = md;
        nxt.resize(1); v.resize(1); val.resize(1); head.resize(mod);
        have_ans = 0;
    }

    void clear() { *this = hash_map_t(mod); }

    bool count(int x) {
        int u = x % mod;
        for(register int i = head[u]; i; i = nxt[i]) {
            if(v[i] == x) {
                have_ans = 1;
                lastv = x;
                lastans = val[i];
                return 1;
            }
        }
        return 0;
    }

    void ins(int x, int y) {
        int u = x % mod;
        nxt.push_back(head[u]); head[u] = ++tot;
        v.push_back(x); val.push_back(y);
    }

    int qry(int x) {
        if(have_ans && lastv == x) return lastans;
        count(x);
        return lastans;
    }
};

const int N = 505;

int a[N], b[N], l[N], r[N];
int n, k, sum, ans;

void solve(int x) {
    if(x < ans) return;
    for(register int i = 1; i <= n; i++) b[i] = a[i] % x;
    sort(b + 1, b + n + 1);
    for(register int i = 1; i <= n; i++) l[i] = l[i - 1] + b[i];
    for(register int i = n; i >= 1; i--) r[i] = r[i + 1] + (b[i] == 0 ? 0 : (x - b[i]));
    if(l[n] == 0) { ans = x; return; }
    for(register int i = 1; i < n; i++) {
        if(l[i] >= r[i + 1] && l[i] <= k) {
            ans = x; return;
        }
    }
}

signed main() {
    read(n); read(k);
    for(register int i = 1; i <= n; i++) read(a[i]), sum += a[i];
    for(register int i = 1; i * i <= sum; i++) {
        if(sum % i == 0) {
            solve(i);
            solve(sum / i);
        }
    }
    print(ans, '\n');
    return 0;
}