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: Accepted)

#include<stdio.h>
int main() {
	long long int n,m, memo = 0;
	long long int a[100005] = {}, sum[100005] = {};
	scanf("%lld", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		sum[i] =sum[i-1]+ a[i];
		memo += i;
	}

	if (sum[n]%memo != 0) { printf("NO"); return 0; }
	m = sum[n] / memo;
	for (int i = 2; i <= n; i++) {
		
		m -= a[i] - a[i - 1] - sum[n] / memo;
		if((a[i]-a[i-1]-sum[n]/memo)%n!=0|| (a[i] - a[i - 1] - sum[n] / memo)>0){ printf("NO"); return 0; }
	}
	if(m<0){ printf("NO"); return 0; }
	printf("YES");
}