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>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define min(a, b) a < b ? a : b
#define max(a, b) a > b ? a : b

#define MAXN 100000

int main(void)
{
	long long i;
	long long N;
	//int ok;
	long long sumn = 0, suma = 0, k;

	//scanf("%lld", &N);

	long long A[MAXN];
	long long d[MAXN];

	scanf("%lld", &N);

	for (i = 0; i < N; i++) {
		scanf("%lld", &A[i]);
		suma += A[i];
	}

	sumn = N * (N + 1) / 2;
	if ((suma % sumn) != 0) {
		//ok = 0;
		printf("NO\n");
	} else {
		k = suma / sumn;
		for (i = 0; i < N; i++) {
			d[i] = A[(i + 1) % N] - A[i] - k;
		}
		for (i = 0; i < N; i++) {
			/*if (d[i] % N == 0 && d[i] <= 0) {
				ok = 1;
			} else {
				ok = 0;
				break;
			}*/
			//ok = 1;
			if (d[i] % N != 0 || d[i] > 0) {
				//ok = 0;
				//break;
				printf("NO\n");
				return (0);
			}
		}
		printf("YES\n");
	}
	/*if (ok) {
		
	} else {
		printf("NO\n");
	}*/
	return (0);
}