A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Consider a permutation p of length n, we build a graph of size n using it as follows: 

  * For every 1 ≤ i ≤ n, find the largest j such that 1 ≤ j < i and p_j > p_i, and add an undirected edge between node i and node j 
  * For every 1 ≤ i ≤ n, find the smallest j such that i < j ≤ n and p_j > p_i, and add an undirected edge between node i and node j 



In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.

For clarity, consider as an example n = 4, and p = [3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).

A permutation p is cyclic if the graph built using p has at least one simple cycle. 

Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 10^9+7.

Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3 ≤ n ≤ 10^6).

Output

Output a single integer 0 ≤ x < 10^9+7, the number of cyclic permutations of length n modulo 10^9+7.

Examples

Input


4


Output


16

Input


583291


Output


135712853

Note

There are 16 cyclic permutations for n = 4. [4,2,1,3] is one such permutation, having a cycle of length four: 4 → 3 → 2 → 1 → 4.

Nodes v_1, v_2, …, v_k form a simple cycle if the following conditions hold: 

  * k ≥ 3. 
  * v_i ≠ v_j for any pair of indices i and j. (1 ≤ i < j ≤ k) 
  * v_i and v_{i+1} share an edge for all i (1 ≤ i < k), and v_1 and v_k share an edge. 
#
#    ------------------------------------------------
#           ____          _     Generatered using
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#         | |    __ _  __| | ___ _ __  ______ _
#         | |   / _` |/ _` |/ _ \ '_ \|_  / _` |
#         | |__| (_| | (_| |  __/ | | |/ / (_| |
#          \____\____|\____|\___|_| |_/___\____|
#
#      GNU Affero General Public License v3.0
#    ------------------------------------------------
#    Author   : prophet
#    Created  : 2020-08-09 15:35:32.267436
#    UUID     : WCZHVUdBb3ie4ALi
#    ------------------------------------------------
#
production = True

import sys, math, collections, bisect, itertools, heapq, decimal, random, copy, re

def input(f = 0, m = 0):

    if m > 0: return [input(f) for i in range(m)]
    else:
        l = sys.stdin.readline()[:-1]

        if f >= 10:
            u = False
            f = int(str(f)[-1])
        else: u = True

        if f == 0: p = [l]
        elif f == 1: p = list(map(int, l.split()))
        elif f == 2: p = list(map(float, l.split()))
        elif f == 3: p = list(l)
        elif f == 4: p = list(map(int, list(l)))
        elif f == 5: p = l.split()

        return p if u else p[0]

def out(l, f = 0, n = True):

    if f == 0: p = str(l)
    elif f == 1: p = " ".join(map(str, l))
    elif f == 2: p = "\n".join(map(str, l))
    elif f == 3: p = "".join(map(str, l))

    print(p, end = "\n" if n else "")

def log(*args):
    if not production:
        print("$$$", end = "")
        print(*args)

enu = enumerate
ter = lambda a, b, c: b if a else c
ceil = lambda a, b: -(-a // b)

def mapl(i, f = 0):

    if f == 0: return list(map(int, i))
    elif f == 1: return list(map(str, i))
    elif f == 2: return list(map(list, i))

#
#   >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#
mb = 10 ** 9 + 7

def solve():

    n = input(11)

    t = 1

    for i in range(n - 1):
        t *= 2
        t %= mb
    
    a = 1
    for i in range(1, n + 1):
        a *= i
        a %= mb

    log(a, t)
    out((a - t) % mb)
    log()
    return


solve()

#
#   >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#
