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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

//#define FILE_IO

typedef long long LL;
int N;
int v[100005];

void yes()
{
    printf("YES\n");
    exit(0);
}

void no()
{
    printf("NO\n");
    exit(0);
}

int main()
{
    #ifdef FILE_IO
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    #endif

    scanf("%d", &N);
    LL sum = 0;
    for(int i = 1; i <= N; i++)
        scanf("%d", &v[i]), sum += v[i];
    LL clr = 1LL * N * (N + 1) / 2;
    if(sum % clr != 0)  no();
    LL moves = sum / clr;
    v[N + 1] = v[1];
    for(int i = 1; i <= N; i++)
    {
        LL dif = v[i + 1] - v[i];
        LL kd = moves - dif;
        if(kd % N != 0 || kd < 0) no();
    }

    yes();
    return 0;
}