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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdint.h>

int a[100000];
int v[100000];
int main(int argc, char *argv[])
{
	int n, i, k, d, p;
	int64_t sum;
	scanf("%d", &n);
	sum = 0;
	for(i = 0; i < n; i ++){
		scanf("%d", &a[i]);
		sum += a[i];
	}
	d = (1 + n) / 2;
	if(sum % d){
		puts("NO");
		return 0;
	}
	d = sum / d;
	d = sum / d;
	p = 0;
	v[0] = (a[n - 1] - a[0] + d) / n;
	p += v[0];
	for(i = 1; i < n; i++){
		v[i] = (a[i - 1] - a[i] + d) / n;
		p += v[i];
	}
	for(i = 0; i < n; i++){
		for(k = 0; k < n; k++){
			a[k] -= v[i] * ((n + k - i) % n + 1);
		}
	}
	for(i = 0; i < n; i++){
		if(a[i]){
			puts("NO");
			return 0;
		}
	}
	puts("YES");
}