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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
long long n, a[100010], k, i, cnt, x, tmp;
int main(){
	puts("YES");
	for(scanf("%lld", &n); i<n; i++){
		scanf("%lld", a+i);
		k+=a[i];
		cnt+=i+1;
	}
	if(k%cnt!=0)	return 0*printf("NO");
	k/=cnt;
	cnt=0;
	for(i=0; i<n; i++){
		tmp=a[i]-a[(i+1)%n];
		if((tmp+k)%n!=0)	return 0*printf("NO");
		x=(tmp+k)/n;
		cnt+=x;
	}
	if(cnt!=k)	return 0*printf("NO");
	puts("YES");
	return 0;
}