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 Python (3.4.3) (judge verdict: Wrong Answer)

# coding: utf-8
num_N = int(input())
l_stone = list(map(int, input().split()))
print(l_stone)

sum_num_N = sum(list(range(1, num_N + 1)))

def solve():
    if sum(l_stone) % sum_num_N != 0:
        return "NO"

    num_times = sum(l_stone) // sum_num_N

    l_dif = []
    for i in range(num_N - 1):
        l_dif.append((l_stone[i + 1] - l_stone[i]))
    l_dif.append(l_stone[0] - l_stone[num_N - 1])

    l_dif_arr = list(map(lambda x: x - num_times, l_dif))

    num_dif_times = 0

    for dif in l_dif_arr:
        if dif == 0:
            continue
        elif dif % num_N == 0:
            num_dif_times += (dif * -1) // num_N
        else:
            break

    if num_dif_times == num_times:
        return "YES"

    return "NO"

print(solve())