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`.

Examples

Input

5
4 5 1 2 3


Output

YES


Input

5
6 9 12 10 8


Output

YES


Input

4
1 2 3 1


Output

NO
#include <cstdio>
int N, a[100001];
long long S, C;
int main()
{
	scanf("%d", &N);
	if (N == 1)
	{
		puts("YES");
		return 0;
	}
	for (int i = 1; i <= N; i++)
	{
		scanf("%d", a + i);
		S += a[i];
	}
	if (S % ((long long)N * (N + 1) / 2))
	{
		puts("NO");
		return 0;
	}
	S /= (long long)N * (N + 1) / 2;
	for (int i = 1; i <= N; i++)
	{
		long long D = S - a[i % N + 1] + a[i];
		if (D < 0 || D % N)
		{
			puts("NO");
			return 0;
		}
		C += D / N;
	}
	puts(S == C ? "YES" : "NO");
	return 0;
}
