Problem:
Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main()
{
    int n, a[100010];
    int flag = 0, b, c;
    int i;
    scanf("%d", &n);
    if (n == 1)
    {
        printf("YES\n");
        return 0;
    }
    scanf("%d", &a[0]);
    scanf("%d", &a[1]);
    b = a[1] - a[0];
    for (i = 2; i < n; i++)
    {
        if (a[i] - a[i - 1] != b && flag == 0)
        {
            flag++;
            c = a[i] - a[i - 1];
        }
        if (a[i] - a[i - 1] != b && a[i] - a[i - 1] != c && flag == 1)
        {
            flag++;
            break;
        }
    }
    if (a[0] - a[n - 1] != b && flag == 0)
    {
        c = a[0] - a[n - 1];
    }
    if (a[0] - a[n - 1] != b && a[0] - a[n - 1] != c && flag == 1)
    {
        flag++;
    }
    if (flag >= 2 || (c - b) * (c - b) != n * n)
    {
        printf("NO\n");
    }
    else
        printf("YES\n");
}