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 N;
int a[100000];
void core() {
	int N;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &a[i]);
	}
	int flag;
	while (1) {
		flag = 0;
		for (int i = 0; i < N; i++) {
			if (a[i] < 0) {
				flag = -1;
				break;
			} else if (a[i] > 0) {
				flag = 1;
			}
		}
		if (flag <= 0) {
			break;
		}

		int t = 0;
		int d = 1;
		for (int i = 1; i < N; i++) {
			if (a[i - 1] > a[i]) {
				t = i;
				d = (a[i - 1] - a[i]) / N + 1;
				break;
			}
		}
		for (int i = 0; i < N; i++) {
			a[(t + i) % N] -= (i + 1) * d;
		}
	}
	if (flag == 0) {
		printf("YES\n");
	} else {
		printf("NO\n");
	}
}

int main() {
	core();
	return 0;
}