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
import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000)

def getN():
    return int(input())
def getList():
    return list(map(int, input().split()))
import math

n = getN()
nums = getList()
if sum(nums) % ((n * (n+1)) // 2) != 0:
    print("NO")
    sys.exit()
n_operate = sum(nums) // ((n * (n+1)) // 2)

v_valid = n_operate
v_sign = n

for i, j in zip(nums, nums[1:] + [nums[0]]):
    diff = j - i
    if diff > v_valid:
        print("NO")
        sys.exit()
    if (v_valid - diff) % v_sign != 0:
        print("NO")
        sys.exit()
    n_operate -= (v_valid - diff) // v_sign
    # print(n_operate)
if n_operate == 0:
    print("YES")
else:
    print("NO")
# print("YES")