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
n=int(input())
arr=list(map(int,input().split()))
if n==1:
  print('YES')
else:
  sum1=sum(arr)
  sum2=(n*(n+1))//2
  moves=sum1//sum2
  cnt=0
  if sum1%sum2!=0:
    print('NO')
  else:
    flag=True
    arr2=[]
    for i in range(n-1):
      tmp=arr[i]+moves
      if arr[i+1]==tmp:
        continue
      else:
        diff=tmp-arr[i+1]
        if diff%n!=0:
          flag=False
        else:
          arr2.append(n-i)
          cnt+=diff//n
          tmp=arr[i+1]-diff//n
          if tmp>n*(moves-diff//n):
            flag=False
    if flag==False:
      print('NO')
    else:
      if len(arr2)==moves:
        if arr[0]==sum(arr2):
          print('YES')
        else:
          print('NO')
      elif len(arr2)>moves:
        print('NO')
      else:
        if 1*(moves-len(arr2))<=arr[0]-sum(arr2)<=n*(moves-len(arr2)):
          print('YES')
        else:
          print('NO')