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: Time Limit Exceeded)

#include <stdio.h>
int main(void){
    int n;
    scanf("%d",&n);
    int a[n];
    int loop1;
    for(loop1=0;loop1<n;loop1++){
        scanf("%d",&a[loop1]);
    }
    //計算量としてはO(n^2)までならいける
    int max=-1;
    int flag,now,total;
    while(1){
        max=0;
        for(loop1=0;loop1<n;loop1++){
            if(a[max]<a[loop1])max=loop1;
        }
        flag=0;
        
        now=max;
        for(loop1=1;loop1<=n;loop1++){
            now++;
            if(now>=n){
                now=0;
            }
            a[now]-=loop1;
            if(a[now]<0){
                flag=1;
                loop1=n+5;
            }
        }
        if(flag){
            printf("NO\n");
            return 0;
        }
        total=0;
        for(loop1=0;loop1<n;loop1++){
            total+=a[loop1];
        }
        if(total==0){
            printf("YES\n");
            return 0;
        }
    }
}