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. 
'''input
4
'''

import math
import sys
import itertools
import random
import bisect
import heapq
import collections

def lcm(x, y):
	return (x*y)//math.gcd(x,y)
def hcf(x, y):
	return math.gcd(x, y)
def primes(n):
	prime=[True for _ in range(0, n+1)]
	res=[]
	p=2
	while p*p<=n:
		if prime[p]==True:
			for i in range(p*p,n+1,p):
				prime[p]=False
		p+=1
	for i in range(0, n+1):
		if prime[i]==True:
			res.append(i)
	return res
def primeFactors(n):
	res=[]
	while n%2==0:
		res.append(2)
		n//=2
	for i in range(3, math.sqrt(n)+1, 2):
		while n%i==0:
			res.append(i)
			n//=i
	if n>2:
		res.append(n)
	return res

def power( x, y, p): 
    res = 1 # Initialize result  
    x = x % p # Update x if it is more than  
              # or equal to p  
    while (y > 0) :  
          
        # If y is odd, multiply x with result  
        if (y & 1) : 
            res = (res * x) % p  
  
        # y must be even now  
        y = y >> 1 # y = y/2  
        x = (x * x) % p  
  
    return res  

def modFact(n, p): 
    if n >= p: 
        return 0    
  
    result = 1
    for i in range(1, n + 1): 
        result = (result * i) % p  
  
    return result 

# print(modFact(3, 10**9+7))

# for _ in range(0, int(input())):
n=int(input())                                 # Single Digit Input
	# n, k = map(int, input().strip().split())    # Many Digit Input
	# l=list(map(int, input().strip().split()))      # List Input
	# s=list(input())                                # Character Array Input

print((modFact(n, (10 ** 9 + 7))-(power(2, n-1, 10 ** 9 + 7)))%(10**9+7))