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
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

def readln(ch):
    _res = list(map(int,str(input()).split(ch)))
    return _res

def pend(x,f,n):
    b = [0 for i in range(0,n+1)]
    for i in range(1,n+1):
        tmp =  f[i] - f[i-1]
        if tmp % n != 0:
            return 'NO'
        b[i] = -(tmp // n)
    b[0] = 0
    k = 0
    l = 0
    for i in range(0,n):
        b[i] = b[i-1] + b[i]
        k = k + n - i
        l = l + (n - i) * b[i]
    x = x - l
    if x < 0 : return 'NO'
    if x % k > 0: return 'NO'
    a[0] = x // k
    for i in range(1,n):
        a[i] = a[0] + b[i]
    for i in range(0,n):
        if(a[i] < 0):
            return 'NO'
    return 'YES'

n = int(input())
a = readln(' ')
f = [0 for i in range(0,n)]
f[0] = a[0]-a[n-1]
for i in range(1,n):
    f[i] = a[i] - a[i-1]

f.append(f[0])
print(pend(a[n-1],f,n))
# an-1 = nx1+n-1x2
