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. 
import sys
from collections import defaultdict as dd
from collections import deque
from functools import *
from fractions import Fraction as f
from copy import *
from bisect import *	
from heapq import *
#from math import *
from itertools import permutations 
 
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def inc(d,c):
	d[c]=d[c]+1 if c in d else 1
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def comp(a,b):
	if(a>b):
		return 2
	return 2 if a==b else 0		
def gi():	
	return [xx for xx in input().split()]
def fi():
	return int(input())
def pro(a): 
	return reduce(lambda a,b:a*b,a)		
def swap(a,i,j): 
	a[i],a[j]=a[j],a[i]	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())	
def gh():
	sys.stdout.flush()
def isvalid(i,j):
	return 0<=i<n and 0<=j<n	
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)
def modInverse(a, m) : 
    m0 = m 
    y = 0
    x = 1
  
    if (m == 1) : 
        return 0
  
    while (a > 1) : 

        q = a // m 
  
        t = m 

        m = a % m 
        a = t 
        t = y 
  

        y = x - q * y 
        x = t 

    if (x < 0) : 
        x = x + m0 
  
    return x 		
def bo(i):
	return ord(i)-ord('A')
def ck(x,y):
	return int(bin(x)[2:]+bin(y)[2:],2)
t=1
mn=10**6+1
mod=10**9+7
jj=modInverse(6,mod)
def nc3(n):
	return ((((n-1)%mod*n%mod)%mod*(n-2))%mod*jj)%mod

fact=[1]*(mn+1)
for i in range(1,mn+1):
	fact[i]=(i*fact[i-1])%mod

while t>0:
	t-=1
	n=fi()
	#print(nc3(n),fact[n-3])
	print((fact[n]-pow(2,n-1,mod)+mod)%mod)