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
# from collections import defaultdict, deque
# import math
# import copy
from bisect import bisect_left, bisect_right
# import heapq

# sys.setrecursionlimit(1000000)

# input aliases
input = sys.stdin.readline

getS = lambda: input().strip()
getN = lambda: int(input())
getList = lambda: list(map(int, input().split()))
getZList = lambda: [int(x) - 1 for x in input().split()]

INF = 10 ** 20
MOD = 1000000007

def solve():
    n = getN()
    nums = getList()

    bulk = n*(n+1)//2
    ope = sum(nums) // bulk
    if sum(nums) % bulk != 0:
        print("NO")
        return
    nums = nums + [nums[0]]

    tmp = 0
    for i, j in zip(nums, nums[1:]):
        if j - i > ope:
            print("NO")
            return
        else:
            expected = i + ope
            if (expected - j) % n != 0:
                print("NO")
                return
            tmp += (expected - j) // n
    # print(tmp)
    if tmp == ope:
        print("YES")
    else:
        print("NO")


def main():
    n = getN()
    for _ in range(n):
        solve()
if __name__ == "__main__":
    # main()
    solve()
