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
# coding:utf-8

import sys
# from collections import Counter, defaultdict

INF = float('inf')
MOD = 10 ** 9 + 7

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return input()


def main():
    n = II()
    A = LI()

    # 1. Aの総和から操作の回数が分かる
    k, mod = divmod(sum(A), (n * (n + 1) // 2))
    if mod:
        return 'NO'

    A.append(A[0])
    diff = []
    for a1, a2 in zip(A[:-1], A[1:]):
        diff.append(a2 - a1)

    # 2. 逆に石を配る操作を考える
    # 一回の操作で1つのdkは-(n - 1), それ以外は+1される

    # 3. {di - k}は負かつnの倍数になっている必要がある
    return 'YES' if not any([(d - k) % n or d > k for d in diff]) else 'NO'


print(main())
