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>
#include <math.h>

int cmp(const void *x, const void *y) {
  return *(int *)x - *(int *)y;
}

int main() {
  int n, k, ans = 1;
  scanf("%d%d", &n ,&k);
  int a[n], sum = 0;
  for (size_t i = 0; i < n; i++) {
    scanf("%d", a+i);
    sum += a[i];
  }
  for (size_t i = 1; i < (int)sqrt((double)sum)+1; i++) {
    if (sum%i == 0) {
      int d1 = i, d2 = sum/i;
      int r1[n], r2[n], r1sum = 0, r2sum = 0, cnt = 0;
      for (size_t j = 0; j < n; j++) {
        r1[j] = a[j] % d1;
        r2[j] = a[j] % d2;
        r1sum += r1[j];
        r2sum += r2[j];
      }
      qsort(r1, n, sizeof(int), cmp);
      qsort(r2, n, sizeof(int), cmp);
      for (size_t j = n-r1sum/d1; j < n; j++) {
        cnt += d1-r1[j];
      }
      ans = (cnt <= k) ? d1 : ans;
      cnt = 0;
      for (size_t j = n-r2sum/d2; j < n; j++) {
        cnt += d2-r2[j];
      }
      if (cnt <= k) {
        ans = d2;
        break;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}