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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AGC010
{
    /*
    class Program1
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            var a = Console.ReadLine().Split(' ').Select(t => int.Parse(t)).ToArray();
            int n_e = a.Count(x => x % 2 == 0);
            int n_o = a.Length - n_e;

            if(n_o % 2 != 0)
            {
                // 奇数が奇数個あると、絶対に最後に奇数が残る
                Console.WriteLine("NO");
            }
            else
            {
                // 偶数個の奇数は、偶数の個数を変えずに削除できる
                // 偶数だけの列は必ず1つまで減らせる
                Console.WriteLine("YES");
            }
        }
        
    }
    */
    class Program2
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            var a = Console.ReadLine().Split(' ').Select(t => int.Parse(t)).ToArray();

            // 1回で取る数の合計
            int take_amount = n * (n - 1) / 2;

            int sum = a.Sum();

            if(sum % take_amount != 0)
            {
                Console.WriteLine("NO");
                return;
            }

            // 取る回数
            int k = sum / take_amount;

            int[] d = new int[n];
            for(int i = 0; i < n - 1; ++i)
            {
                d[i] = a[i + 1] - a[i];
            }
            d[n - 1] = a[0] - a[n - 1];

            if (d.Select(x => x - k).All(x => x % n == 0))
            {
                Console.WriteLine("YES");
                return;
            }
            Console.WriteLine("NO");
        }

    }
}