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 find_max(int a[], int n)
{
	int max_index = 0;
	int i;
	
	for (i = 1; i < n; i++) {
		if (a[max_index] < a[i]) {
			max_index = i;
		}
	}
	
	return max_index;
}

int main(void)
{
	int i;

	int n;
	scanf("%d", &n);
	
	int a[n];
	for (i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	
	for (;;) {
		int index = find_max(a, n);
		if (a[index] == 0) {
			printf("YES\n");
			return 0;
		}
		for (i = 1; i <= n; i++) {
			a[(index + i) % n] -= i;
			if (a[(index + i) % n] < 0) {
				printf("NO\n");
				return 0;
			}
		}
	}
}