There are N blocks arranged in a row, numbered 1 to N from left to right. Each block has a weight, and the weight of Block i is A_i. Snuke will perform the following operation on these blocks N times:

* Choose one block that is still not removed, and remove it. The cost of this operation is the sum of the weights of the blocks that are connected to the block being removed (including itself). Here, two blocks x and y ( x \leq y ) are connected when, for all z ( x \leq z \leq y ), Block z is still not removed.



There are N! possible orders in which Snuke removes the blocks. For all of those N! orders, find the total cost of the N operations, and calculate the sum of those N! total costs. As the answer can be extremely large, compute the sum modulo 10^9+7.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

For all of the N! orders, find the total cost of the N operations, and print the sum of those N! total costs, modulo 10^9+7.

Examples

Input

2
1 2


Output

9


Input

4
1 1 1 1


Output

212


Input

10
1 2 4 8 16 32 64 128 256 512


Output

880971923
def examA():
    N = DI()/dec(7)
    ans = N
    print(N)
    return

def examB():
    class combination():
        # 素数のmod取るときのみ　速い
        def __init__(self, n, mod):
            self.n = n
            self.mod = mod
            self.fac = [1] * (n + 1)
            self.inv = [1] * (n + 1)
            for j in range(1, n + 1):
                self.fac[j] = self.fac[j - 1] * j % mod

            self.inv[n] = pow(self.fac[n], mod - 2, mod)
            for j in range(n - 1, -1, -1):
                self.inv[j] = self.inv[j + 1] * (j + 1) % mod

        def comb(self, n, r):
            if r > n or n < 0 or r < 0:
                return 0
            return self.fac[n] * self.inv[n - r] * self.inv[r] % self.mod
    N = I()
    A = LI()
    C = combination(N,mod)
    ans = 0
    cnt = 0
    for i in range(N):
        cnt += C.fac[N]*pow(i+1,mod-2,mod)
        cnt %= mod
        #print(cnt)
    #print(cnt)
    ans += cnt*A[0]
    for i in range(1,N):
        cnt = cnt + C.fac[N]*pow(i+1,mod-2,mod) - C.fac[N]*pow(N-i+1,mod-2,mod)
        cnt += mod
        cnt %= mod
        a = A[i]
        ans += cnt*a
        ans %= mod
        #print(cnt)
    print(ans)
    return

def examC():
    ans = 0
    print(ans)
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examB()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""