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. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

import math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
from collections import defaultdict
MOD = 1000000007
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
	
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
		
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def da():
	n, m = map(int, input().split())
	a = list(map(int, input().split()))
	return n,m, a 
 
 
def dva():
	
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
def lol(lst,k):
	k=k%len(lst)
	ret=[0]*len(lst)
	for i in range(len(lst)):
		if i+k<len(lst) and i+k>=0:
			ret[i]=lst[i+k]
		if i+k>=len(lst):
			ret[i]=lst[i+k-len(lst)]
		if i+k<0:
			ret[i]=lst[i+k+len(lst)]
	return(ret)
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m 
 
 
def fact(n):
	tc = []
	ans = {}
	d = 2
	while d * d <= n:
		if n % d == 0:
			tc.append(d)
			n //= d
		else:
			d += 1
	if n > 1:
		tc.append(n)
	for i in tc:
		ans[i] = ans.get(i, 0) + 1
	return ans


def to_zero(n, s):
	lst = s[0]
	ans = []
	for i in range(n):
		if lst == s[i]:
			continue
		else:
			if lst == '0':
				lst = '1'
				ans.append(i)
			else:
				lst = '0'
				ans.append(i)
	if lst == '1':
		ans.append(n)
	return ans


n = int(input())
r = 1
for i in range(1, n + 1):
	r *= i
	r %= MOD
l = 1
for i in range(1, n):
	l *= 2
	l %= MOD
r -= l
print(r)
