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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

int compare_int(const void *a,const void *b){
	return *(long long*)a-*(long long*)b;
}

int main(){
	int n,j;
	long long a[520]={0},k,i;
	scanf("%d%lld",&n,&k);
	for(i=0;i<n;i++)scanf("%lld",&a[i+1]);n++;
	qsort(a,n,sizeof(long long),compare_int);
	long long sum=0;
	for(i=0;i<n;i++)sum+=a[i];
	if(sum<=k){
		printf("%lld\n",sum);
		return 0;
	}
	long long yak[5000]={1},c=1;
	for(i=2;i*i<=sum;i++){
		if(sum%i==0)yak[c++]=i;
	}
	for(i=0;i<c;i++)yak[c+i]=sum/yak[i];
	if(yak[c-1]*yak[c-1]==sum)c=2*c-1;
	else c=2*c;
	long long ans=1;
	for(i=1;i<c;i++){
		long long y=yak[i],tmp=0,tmm=0,tmh=0,tmhm=0;
		long long pa[520];
		for(j=0;j<n;j++){
			pa[j]=a[j]%y;
			if(pa[j]>y-pa[j])pa[j]=y-pa[j];
			else pa[j]= -pa[j];
			if(pa[j]<0){
				tmhm-=pa[j];
			}
			else {
				tmh+=pa[j];
			}
		}
		qsort(pa,n,sizeof(long long),compare_int);
		while(tmh>tmhm){
			for(j=n-1;j>=0;j--){
				tmhm+=(y-pa[j]);
				tmh-=(pa[j]);
				if(tmh==tmhm)break;
			}
		}while(tmh<tmhm){
			for(j=0;j<n;j++){
				tmh+=(y+pa[j]);
				tmhm+=(pa[j]);
				if(tmh==tmhm)break;
			}
		}
		if(tmh<=k)ans=max(ans,y);
	}
	printf("%lld\n",ans);
	return 0;
}