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()
{
    long double n;
    long double flag = 0, b, c;
    long double flag2 = 0;
    long double i;
    long double sum;
    long double tmp;
    long double x0, x1, xa, xb;
    scanf("%Lf", &n);
    if (n <= 1)
    {
        printf("YES\n");
        return 0;
    }
    scanf("%Lf", &xa);
    scanf("%Lf", &xb);
    b = xb - xa;
    tmp = xb + xa;
    x0 = xa;
    xa = xb;
    for (i = 2; i < n; i++)
    {
        scanf("%Lf", &xb);
        if (xb - xa != b && flag == 0)
        {
            flag++;
            c = xb - xa;
        }
        if (xb - xa != b && xb - xa != c && flag == 1)
        {
            flag++;
            break;
        }
        tmp += xb;
        xa = xb;
    }
    if (x0 - xa != b && flag == 0)
    {
        c = x0 - xa;
    }
    if (x0 - xa != b && x0 - xa != c && flag == 1)
    {
        flag++;
    }
    if (b < c)
    {
        sum = c;
        c = b;
        b = sum;
    }
    if (flag >= 2 || b - c != n || tmp != n * (n + 1) / 2 * b)
    {
        printf("NO\n");
    }
    else
        printf("YES\n");
}