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())
A=list(map(int,input().split()))
N=((n)*(n+1))//2

if sum(A)%N!=0:
    print("NO")
    exit()
elif n==1:
    print("YES")
elif n==2:
    if sum(A)%3==0 and max(A)<=2*min(A):
        print("YES")
    else:
        print("NO")
else:
    T=(sum(A)//N)
    A+=[A[0]]
    cnt=0
    for i in range(1,len(A)):
        if (A[i]-A[i-1])==T:
            #print(A[i],A[i-1])
            continue
        else:
            if A[i-1]>A[i]-T:
                B=((A[i-1]-A[i])//n)+1
                cnt+=B
                #print(A[i],A[i-1],n*B)
                if (A[i]-A[i-1]+n*B)==T:
                    continue
                else:
                    print("NO")
                    exit()
            else:
                print("NO")
                exit()
    if T==cnt:
        print("YES")
    else:
        print("NO")
    
