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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define REP(i, n) for(int i=0; i<(n); ++i)
#define FOR(i, a, b) for(int i=(a); i<(b); ++i)
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)

#define DEBUG(x) cout<<#x<<": "<<(x)<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";REP(i, v.size())cout<<' '<<v[i];cout<<'\n'
#define ALL(a) (a).begin(), (a).end()

template<typename T> inline void CHMAX(T& a, const T b) {if(a<b) a=b;}
template<typename T> inline void CHMIN(T& a, const T b) {if(a>b) a=b;}

constexpr ll MOD=1000000007ll;
// constexpr ll MOD=998244353ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

const double EPS=1e-11;
#define EQ0(x) (abs((x))<EPS)
#define EQ(a, b) (abs((a)-(b))<EPS)

// 約数を求める
vl divisor(ll n){
	vl res;
	for(ll i=1; i*i<=n; ++i){
		if(n%i==0){
			res.push_back(i);
			if(i!=n/i){
				res.push_back(n/i);
			}
		}
	}
	return res;
}

const ll INF=1e18;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	// cout<<setprecision(10)<<fixed;
	int n;
	ll k, a[525];
	cin>>n>>k;
	ll sum=0, ma=0;
	REP(i, n){
		cin>>a[i];
		sum+=a[i];
		CHMAX(ma, a[i]);
	}
	sort(a, a+n);
	vl v=divisor(sum);
	ll ans=1;
	for(auto &x : v){
		if(x==1){
			continue;
		}
		if(x==sum){
			if(sum-ma<=k){
				ans=x;
			}
			continue;
		}
		vl b;
		REP(i, n){
			if(a[i]%x==0){
				continue;
			}
			b.push_back(a[i]%x);
		}
		sort(ALL(b));
		int m=b.size();
		ll sum[2][525];
		sum[0][0]=0;
		REP(i, m){
			sum[0][i+1]=sum[0][i]+b[i];
		}
		sum[1][m]=0;
		FORR(i, 0, m){
			sum[1][i]=sum[1][i+1]+x-b[i];
		}
		ll tmp=INF;
		REP(i, n+1){
			if(abs(sum[0][i]-sum[1][i])%x==0){
				CHMIN(tmp, max(sum[0][i], sum[1][i]));
			}
		}
		if(tmp<=k){
			CHMAX(ans, x);
		}
	}
	cout<<ans<<'\n';
	return 0;
}