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
def solve(ar):
  n = len(ar)
  if sum(ar) % (n * (n+1) // 2) != 0:
    return False
  l = sum(ar) // (n * (n+1) // 2)
  ar2 = []
  for i in range(n-1):
    if i == 0:
      ar2.append(ar[0])
    else:
      ar2.append(ar[i] - ar[i-1] - l)

  b = [0]
  for i in range(1, n-1):
    if ar2[i] % n != 0:
      return False
    bi = - ar2[i] // n
    if bi < 0:
      return False
    b.append(bi)

  a = ar[0]
  a2 = l
  for i in range(1, n-1):
    a -= b[i] * (n-i+1)
    a2 -= b[i]

  bn = a - a2
  b0 = a2 - bn
  if bn < 0 or b0 < 0:
    return False

  return True

n = int(input())
ar = [int(x) for x in input().split()]
if solve(ar):
  print("YES")
else:
  print("NO")



