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. 
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))
sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d


def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_right(a, x)
    if i != len(a):
        return i
    else:            
        return -1

def main():
    try:
        n=I()
        ans=1
        for x in range(1,n+1):
            ans=((ans%P)*(x%P))%P
        
        temp=2
        for x in range(n-2):
            temp=((temp%P)*(2%P))%P
        
        print((ans%P-temp%P)%P)

    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    #for _ in range(I()):main()
    for _ in range(1):main()


